STAT1029 - Lecture 7 Supplementary Notes

Dr. Erwin George, School of Computing and Mathematical Sciences, University of Greenwich

September 19, 2012

Dr. Erwin George, Dept. of Mathematics

STAT1029

TIPS Invest time learning the language of Mathematics/Statistics/Computing (with all of its special cases and exceptions and conventions).

Dr. Erwin George, Dept. of Mathematics

STAT1029

TIPS Invest time learning the language of Mathematics/Statistics/Computing (with all of its special cases and exceptions and conventions). Review constantly.

Dr. Erwin George, Dept. of Mathematics

STAT1029

TIPS Invest time learning the language of Mathematics/Statistics/Computing (with all of its special cases and exceptions and conventions). Review constantly. Do assignments, tutorials, etc. Practise, practise, practise.

Dr. Erwin George, Dept. of Mathematics

STAT1029

TIPS Invest time learning the language of Mathematics/Statistics/Computing (with all of its special cases and exceptions and conventions). Review constantly. Do assignments, tutorials, etc. Practise, practise, practise. Read lecture notes and try to learn from lectures. Ideally read the lecture material on a topic before the relevant lecture.

Dr. Erwin George, Dept. of Mathematics

STAT1029

TIPS Invest time learning the language of Mathematics/Statistics/Computing (with all of its special cases and exceptions and conventions). Review constantly. Do assignments, tutorials, etc. Practise, practise, practise. Read lecture notes and try to learn from lectures. Ideally read the lecture material on a topic before the relevant lecture. Lectures will not simply be a repeat of lecture notes. They will focus on showing a wider range of examples and highlighting key points. You will have to do significant work outside of the classroom to master the material.

Dr. Erwin George, Dept. of Mathematics

STAT1029

TIPS Invest time learning the language of Mathematics/Statistics/Computing (with all of its special cases and exceptions and conventions). Review constantly. Do assignments, tutorials, etc. Practise, practise, practise. Read lecture notes and try to learn from lectures. Ideally read the lecture material on a topic before the relevant lecture. Lectures will not simply be a repeat of lecture notes. They will focus on showing a wider range of examples and highlighting key points. You will have to do significant work outside of the classroom to master the material. Attend lectures and tutorials. This represents the most efficient way to learn the material.

Dr. Erwin George, Dept. of Mathematics

STAT1029

TIPS Invest time learning the language of Mathematics/Statistics/Computing (with all of its special cases and exceptions and conventions). Review constantly. Do assignments, tutorials, etc. Practise, practise, practise. Read lecture notes and try to learn from lectures. Ideally read the lecture material on a topic before the relevant lecture. Lectures will not simply be a repeat of lecture notes. They will focus on showing a wider range of examples and highlighting key points. You will have to do significant work outside of the classroom to master the material. Attend lectures and tutorials. This represents the most efficient way to learn the material. In summary: KEEP UP. This will be a very fast-paced class and falling behind is very unwise. Mathematics/Statistics is very hierarchical and you generally progress only by first knowing well what went before.
Dr. Erwin George, Dept. of Mathematics STAT1029

→ Much of what we will do is based on the following rules for computing probabilities of compound events (i.e. cases in which we combine events with the words AND, OR, or NOT):

Dr. Erwin George, Dept. of Mathematics

STAT1029

→ Much of what we will do is based on the following rules for computing probabilities of compound events (i.e. cases in which we combine events with the words AND, OR, or NOT): If A and B are events, then

Dr. Erwin George, Dept. of Mathematics

STAT1029

→ Much of what we will do is based on the following rules for computing probabilities of compound events (i.e. cases in which we combine events with the words AND, OR, or NOT): If A and B are events, then
1. P(not A) = 1 − P(A).

Dr. Erwin George, Dept. of Mathematics

STAT1029

→ Much of what we will do is based on the following rules for computing probabilities of compound events (i.e. cases in which we combine events with the words AND, OR, or NOT): If A and B are events, then
1. P(not A) = 1 − P(A). 2. P(A or B [or both]) = P(A) + P(B) − P(A and B).

Dr. Erwin George, Dept. of Mathematics

STAT1029

→ Much of what we will do is based on the following rules for computing probabilities of compound events (i.e. cases in which we combine events with the words AND, OR, or NOT): If A and B are events, then
1. P(not A) = 1 − P(A). 2. P(A or B [or both]) = P(A) + P(B) − P(A and B). 3. P(A and B) = P(A)P(B), provided A and B are independent events (you can largely assume that in STAT1029).

Dr. Erwin George, Dept. of Mathematics

STAT1029

→ Much of what we will do is based on the following rules for computing probabilities of compound events (i.e. cases in which we combine events with the words AND, OR, or NOT): If A and B are events, then
1. P(not A) = 1 − P(A). 2. P(A or B [or both]) = P(A) + P(B) − P(A and B). 3. P(A and B) = P(A)P(B), provided A and B are independent events (you can largely assume that in STAT1029).

→ NOTE 1: in number (2) above, if the events A and B are mutually exclusive (meaning that they CANNOT both happen), then since P(A and B) = 0, the formula just becomes P(A or B) = P(A) + P(B).

Dr. Erwin George, Dept. of Mathematics

STAT1029

→ Much of what we will do is based on the following rules for computing probabilities of compound events (i.e. cases in which we combine events with the words AND, OR, or NOT): If A and B are events, then
1. P(not A) = 1 − P(A). 2. P(A or B [or both]) = P(A) + P(B) − P(A and B). 3. P(A and B) = P(A)P(B), provided A and B are independent events (you can largely assume that in STAT1029).

→ NOTE 1: in number (2) above, if the events A and B are mutually exclusive (meaning that they CANNOT both happen), then since P(A and B) = 0, the formula just becomes P(A or B) = P(A) + P(B). It will often be the case that events are mutually exclusive.

Dr. Erwin George, Dept. of Mathematics

STAT1029

→ Much of what we will do is based on the following rules for computing probabilities of compound events (i.e. cases in which we combine events with the words AND, OR, or NOT): If A and B are events, then
1. P(not A) = 1 − P(A). 2. P(A or B [or both]) = P(A) + P(B) − P(A and B). 3. P(A and B) = P(A)P(B), provided A and B are independent events (you can largely assume that in STAT1029).

→ NOTE 1: in number (2) above, if the events A and B are mutually exclusive (meaning that they CANNOT both happen), then since P(A and B) = 0, the formula just becomes P(A or B) = P(A) + P(B). It will often be the case that events are mutually exclusive. → NOTE 2: Results (2) and (3) above generalise in the natural way to more than 2 events.

Dr. Erwin George, Dept. of Mathematics

STAT1029

→ Much of what we will do is based on the following rules for computing probabilities of compound events (i.e. cases in which we combine events with the words AND, OR, or NOT): If A and B are events, then
1. P(not A) = 1 − P(A). 2. P(A or B [or both]) = P(A) + P(B) − P(A and B). 3. P(A and B) = P(A)P(B), provided A and B are independent events (you can largely assume that in STAT1029).

→ NOTE 1: in number (2) above, if the events A and B are mutually exclusive (meaning that they CANNOT both happen), then since P(A and B) = 0, the formula just becomes P(A or B) = P(A) + P(B). It will often be the case that events are mutually exclusive. → NOTE 2: Results (2) and (3) above generalise in the natural way to more than 2 events. For example, if events A, B, and C are independent, then P(A and B and C ) = P(A)P(B)P(C ).

Dr. Erwin George, Dept. of Mathematics

STAT1029

→ Much of what we will do is based on the following rules for computing probabilities of compound events (i.e. cases in which we combine events with the words AND, OR, or NOT): If A and B are events, then
1. P(not A) = 1 − P(A). 2. P(A or B [or both]) = P(A) + P(B) − P(A and B). 3. P(A and B) = P(A)P(B), provided A and B are independent events (you can largely assume that in STAT1029).

→ NOTE 1: in number (2) above, if the events A and B are mutually exclusive (meaning that they CANNOT both happen), then since P(A and B) = 0, the formula just becomes P(A or B) = P(A) + P(B). It will often be the case that events are mutually exclusive. → NOTE 2: Results (2) and (3) above generalise in the natural way to more than 2 events. For example, if events A, B, and C are independent, then P(A and B and C ) = P(A)P(B)P(C ). Similarly, if A, B, and C are mutually exclusive, then P(A or B or C ) = P(A) + P(B) + P(C ).
Dr. Erwin George, Dept. of Mathematics STAT1029

Sampling Without Replacement - Additional Example → EXAMPLE 1 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. What are the following probabilities:

Dr. Erwin George, Dept. of Mathematics

STAT1029

Sampling Without Replacement - Additional Example → EXAMPLE 1 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. What are the following probabilities:
(a) All 3 balls drawn are red (c) Either 3 red or 3 yellow balls are drawn (e) At least one of the balls drawn is yellow (b) All 3 balls drawn are yellow (d) One of the balls drawn is yellow

Dr. Erwin George, Dept. of Mathematics

STAT1029

Sampling Without Replacement - Additional Example → EXAMPLE 1 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. What are the following probabilities:
(a) All 3 balls drawn are red (c) Either 3 red or 3 yellow balls are drawn (e) At least one of the balls drawn is yellow (a) We want P(All 3 balls drawn are red) = (b) All 3 balls drawn are yellow (d) One of the balls drawn is yellow

Dr. Erwin George, Dept. of Mathematics

STAT1029

Sampling Without Replacement - Additional Example → EXAMPLE 1 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. What are the following probabilities:
(a) All 3 balls drawn are red (c) Either 3 red or 3 yellow balls are drawn (e) At least one of the balls drawn is yellow (a) We want P(All 3 balls drawn are red) = P(first ball red AND second ball red AND third ball red) = (b) All 3 balls drawn are yellow (d) One of the balls drawn is yellow

Dr. Erwin George, Dept. of Mathematics

STAT1029

Sampling Without Replacement - Additional Example → EXAMPLE 1 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. What are the following probabilities:
(a) All 3 balls drawn are red (c) Either 3 red or 3 yellow balls are drawn (e) At least one of the balls drawn is yellow (a) We want P(All 3 balls drawn are red) = P(first ball red AND second ball red AND third ball red) = P(first ball red) ∗ P(second ball red) ∗ P(third ball red) = (b) All 3 balls drawn are yellow (d) One of the balls drawn is yellow

Dr. Erwin George, Dept. of Mathematics

STAT1029

Sampling Without Replacement - Additional Example → EXAMPLE 1 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. What are the following probabilities:
(a) All 3 balls drawn are red (c) Either 3 red or 3 yellow balls are drawn (e) At least one of the balls drawn is yellow (a) We want P(All 3 balls drawn are red) = P(first ball red AND second ball red AND third ball red) = P(first ball red) ∗ P(second ball red) ∗ P(third ball red) = 8 7 6 8 = ... = = 0.123077. 15 14 13 65 (b) All 3 balls drawn are yellow (d) One of the balls drawn is yellow

Dr. Erwin George, Dept. of Mathematics

STAT1029

Sampling Without Replacement - Additional Example → EXAMPLE 1 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. What are the following probabilities:
(a) All 3 balls drawn are red (c) Either 3 red or 3 yellow balls are drawn (e) At least one of the balls drawn is yellow (a) We want P(All 3 balls drawn are red) = P(first ball red AND second ball red AND third ball red) = P(first ball red) ∗ P(second ball red) ∗ P(third ball red) = 8 7 6 8 = ... = = 0.123077. 15 14 13 65 (b) We want P(All 3 balls drawn are yellow) = (b) All 3 balls drawn are yellow (d) One of the balls drawn is yellow

Dr. Erwin George, Dept. of Mathematics

STAT1029

Sampling Without Replacement - Additional Example → EXAMPLE 1 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. What are the following probabilities:
(a) All 3 balls drawn are red (c) Either 3 red or 3 yellow balls are drawn (e) At least one of the balls drawn is yellow (a) We want P(All 3 balls drawn are red) = P(first ball red AND second ball red AND third ball red) = P(first ball red) ∗ P(second ball red) ∗ P(third ball red) = 8 7 6 8 = ... = = 0.123077. 15 14 13 65 (b) We want P(All 3 balls drawn are yellow) = P(first ball yellow AND second ball yellow AND third ball yellow) = (b) All 3 balls drawn are yellow (d) One of the balls drawn is yellow

Dr. Erwin George, Dept. of Mathematics

STAT1029

Sampling Without Replacement - Additional Example → EXAMPLE 1 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. What are the following probabilities:
(a) All 3 balls drawn are red (c) Either 3 red or 3 yellow balls are drawn (e) At least one of the balls drawn is yellow (a) We want P(All 3 balls drawn are red) = P(first ball red AND second ball red AND third ball red) = P(first ball red) ∗ P(second ball red) ∗ P(third ball red) = 8 7 6 8 = ... = = 0.123077. 15 14 13 65 (b) We want P(All 3 balls drawn are yellow) = P(first ball yellow AND second ball yellow AND third ball yellow) = P(first ball yellow) ∗ P(second ball yellow) ∗ P(third ball yellow) = (b) All 3 balls drawn are yellow (d) One of the balls drawn is yellow

Dr. Erwin George, Dept. of Mathematics

STAT1029

Sampling Without Replacement - Additional Example → EXAMPLE 1 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. What are the following probabilities:
(a) All 3 balls drawn are red (c) Either 3 red or 3 yellow balls are drawn (e) At least one of the balls drawn is yellow (a) We want P(All 3 balls drawn are red) = P(first ball red AND second ball red AND third ball red) = P(first ball red) ∗ P(second ball red) ∗ P(third ball red) = 8 7 6 8 = ... = = 0.123077. 15 14 13 65 (b) We want P(All 3 balls drawn are yellow) = P(first ball yellow AND second ball yellow AND third ball yellow) = P(first ball yellow) ∗ P(second ball yellow) ∗ P(third ball yellow) = 7 6 5 1 = ... = = 0.076923. 15 14 13 13
Dr. Erwin George, Dept. of Mathematics STAT1029

(b) All 3 balls drawn are yellow (d) One of the balls drawn is yellow

REMINDER : 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. (c) P(3 yellow or 3 red balls drawn) =

Dr. Erwin George, Dept. of Mathematics

STAT1029

REMINDER : 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. (c) P(3 yellow or 3 red balls drawn) = P(3 yellow balls drawn)+P(3 red balls drawn)−P(3 yellow AND 3 red balls drawn) =

Dr. Erwin George, Dept. of Mathematics

STAT1029

REMINDER : 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. (c) P(3 yellow or 3 red balls drawn) = P(3 yellow balls drawn)+P(3 red balls drawn)−P(3 yellow AND 3 red balls drawn) = 8 1 + −0= 65 13

Dr. Erwin George, Dept. of Mathematics

STAT1029

REMINDER : 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. (c) P(3 yellow or 3 red balls drawn) = P(3 yellow balls drawn)+P(3 red balls drawn)−P(3 yellow AND 3 red balls drawn) = 8 5 13 8 1 + = = + −0= 65 65 65 65 13

Dr. Erwin George, Dept. of Mathematics

STAT1029

REMINDER : 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. (c) P(3 yellow or 3 red balls drawn) = P(3 yellow balls drawn)+P(3 red balls drawn)−P(3 yellow AND 3 red balls drawn) = 8 5 13 1 8 1 + = = or 0.2. + −0= 65 65 65 5 65 13

Dr. Erwin George, Dept. of Mathematics

STAT1029

REMINDER : 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. (c) P(3 yellow or 3 red balls drawn) = P(3 yellow balls drawn)+P(3 red balls drawn)−P(3 yellow AND 3 red balls drawn) = 8 5 13 1 8 1 + = = or 0.2. + −0= 65 65 65 5 65 13 (d) P(one ball drawn is yellow) =

Dr. Erwin George, Dept. of Mathematics

STAT1029

REMINDER : 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. (c) P(3 yellow or 3 red balls drawn) = P(3 yellow balls drawn)+P(3 red balls drawn)−P(3 yellow AND 3 red balls drawn) = 8 5 13 1 8 1 + = = or 0.2. + −0= 65 65 65 5 65 13 (d) P(one ball drawn is yellow) = P({1st ball yellow AND next 2 red} OR {1st ball red, AND second yellow, AND third red} OR {1st 2 balls red AND last yellow}).

Dr. Erwin George, Dept. of Mathematics

STAT1029

REMINDER : 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. (c) P(3 yellow or 3 red balls drawn) = P(3 yellow balls drawn)+P(3 red balls drawn)−P(3 yellow AND 3 red balls drawn) = 8 5 13 1 8 1 + = = or 0.2. + −0= 65 65 65 5 65 13 (d) P(one ball drawn is yellow) = P({1st ball yellow AND next 2 red} OR {1st ball red, AND second yellow, AND third red} OR {1st 2 balls red AND last yellow}). Since the events separated by the word OR above are mutually exclusive, this probability is simply the sum of the individual probabilities: P({1st ball yellow AND next 2 red})+ P({1st ball red, AND second yellow, AND third red})+ P({1st 2 balls red AND last yellow}) =

Dr. Erwin George, Dept. of Mathematics

STAT1029

REMINDER : 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. (c) P(3 yellow or 3 red balls drawn) = P(3 yellow balls drawn)+P(3 red balls drawn)−P(3 yellow AND 3 red balls drawn) = 8 5 13 1 8 1 + = = or 0.2. + −0= 65 65 65 5 65 13 (d) P(one ball drawn is yellow) = P({1st ball yellow AND next 2 red} OR {1st ball red, AND second yellow, AND third red} OR {1st 2 balls red AND last yellow}). Since the events separated by the word OR above are mutually exclusive, this probability is simply the sum of the individual probabilities: P({1st ball yellow AND next 2 red})+ P({1st ball red, AND second yellow, AND third red})+ P({1st 2 balls red AND last yellow}) = 7 8 7 8 7 7 8 7 7 + + = 15 14 13 15 14 13 15 14 13
Dr. Erwin George, Dept. of Mathematics STAT1029

REMINDER : 3 balls are drawn at random from a bag containing 8 red balls and 7 yellow balls. (c) P(3 yellow or 3 red balls drawn) = P(3 yellow balls drawn)+P(3 red balls drawn)−P(3 yellow AND 3 red balls drawn) = 8 5 13 1 8 1 + = = or 0.2. + −0= 65 65 65 5 65 13 (d) P(one ball drawn is yellow) = P({1st ball yellow AND next 2 red} OR {1st ball red, AND second yellow, AND third red} OR {1st 2 balls red AND last yellow}). Since the events separated by the word OR above are mutually exclusive, this probability is simply the sum of the individual probabilities: P({1st ball yellow AND next 2 red})+ P({1st ball red, AND second yellow, AND third red})+ P({1st 2 balls red AND last yellow}) = 7 8 7 8 7 7 8 7 7 28 + + = ... = or 0.430769. 15 14 13 15 14 13 15 14 13 65
Dr. Erwin George, Dept. of Mathematics STAT1029

(e) Since the negation of “at least one of the balls drawn is yellow” is “no ball drawn is yellow”, then the easy way to calculate P(at least one of the balls drawn is yellow) is as

Dr. Erwin George, Dept. of Mathematics

STAT1029

(e) Since the negation of “at least one of the balls drawn is yellow” is “no ball drawn is yellow”, then the easy way to calculate P(at least one of the balls drawn is yellow) is as = 1−P(no ball drawn is yellow) =

Dr. Erwin George, Dept. of Mathematics

STAT1029

(e) Since the negation of “at least one of the balls drawn is yellow” is “no ball drawn is yellow”, then the easy way to calculate P(at least one of the balls drawn is yellow) is as = 1−P(no ball drawn is yellow) = 1−P(all 3 balls drawn are red) =

Dr. Erwin George, Dept. of Mathematics

STAT1029

(e) Since the negation of “at least one of the balls drawn is yellow” is “no ball drawn is yellow”, then the easy way to calculate P(at least one of the balls drawn is yellow) is as = 1−P(no ball drawn is yellow) = 1−P(all 3 balls drawn are red) = (from part (a)) 1 − 8 = 65

Dr. Erwin George, Dept. of Mathematics

STAT1029

(e) Since the negation of “at least one of the balls drawn is yellow” is “no ball drawn is yellow”, then the easy way to calculate P(at least one of the balls drawn is yellow) is as = 1−P(no ball drawn is yellow) = 1−P(all 3 balls drawn are red) = (from part (a)) 1 − 57 8 or 0.87692. = 65 65

Dr. Erwin George, Dept. of Mathematics

STAT1029

(e) Since the negation of “at least one of the balls drawn is yellow” is “no ball drawn is yellow”, then the easy way to calculate P(at least one of the balls drawn is yellow) is as = 1−P(no ball drawn is yellow) = 1−P(all 3 balls drawn are red) = (from part (a)) 1 − 57 8 or 0.87692. = 65 65

OPTIONAL: Here is the long way of computing the answer to part (e):
P(at least one of the balls drawn is yellow) = P(exactly one ball yellow OR exactly two balls yellow OR exactly three balls yellow) = P(exactly one ball yellow) + P(exactly two balls yellow) + P(all three balls yellow) ⇒ P(at least one of the balls drawn is yellow) = 28 65 +P(exactly two balls yellow)+ 1 13 (using results from parts (d) and (b)). (1) So the only probability left to calculate is P(exactly two balls yellow). This is possible if A : the first two balls are yellow and the last is red OR B : the first ball is yellow, the next red, and the last yellow, OR C : the 7 6 8 7 8 6 8 7 6 first ball is red and the next two are yellow. The associated probabilities are 15 14 13 , 15 14 13 , and 15 14 13 7 6 8 7 8 6 8 7 6 24 respectively. So P(exactly two balls yellow) = 15 14 13 + 15 14 13 + 15 14 13 = . . . = 65 . So putting this into Equation (1) above, we get P(at least one of the balls drawn is yellow) = 28 65 + 24 65 + 1 13 = 28 65 + 24 65 + 5 65 = 57 65 as expected.

Dr. Erwin George, Dept. of Mathematics

STAT1029

Difference Between Discrete and Continuous Probability Distributions → A discrete probability distribution is a list (or chart or graph) which contains ALL possible outcomes of an experiment along with the probability of each of those outcomes. E.g., An experiment consisting of counting the number of students who attend STAT1029 lecture each week would result in a discrete probability distribution (the number of possible outcomes would be an integer 0 or 1 or 2 or 3 or . . . or 42 (the current number of registered students). → In a continuous probability distribution it is impossible to list all possible outcomes of the experiment. One can give the range of possible outcomes (for example, from 1 to 10 and any real number in between) and draw a graph indicating the frequency/popularity of certain ranges of outcomes. E.g., the height of students at the University of Greenwich during any academic year is a continuous random variable . . . in principle, it could take on any value in a certain range (to be liberal, say from 0.2 metres to say 3 metres).
Dr. Erwin George, Dept. of Mathematics STAT1029

Poisson Probability Distribution - Additional Example → EXAMPLE 2 Suppose that the weekly average number of accidents on a stretch of motorway is 5. (a) What is the probability there will be 2 accidents next week? (b) What is the probability there will be more than 4 accidents next week?

Dr. Erwin George, Dept. of Mathematics

STAT1029

Poisson Probability Distribution - Additional Example → EXAMPLE 2 Suppose that the weekly average number of accidents on a stretch of motorway is 5. (a) What is the probability there will be 2 accidents next week? (b) What is the probability there will be more than 4 accidents next week? (a) We want P(X = 2) where X = “the number of accidents next week”.

Dr. Erwin George, Dept. of Mathematics

STAT1029

Poisson Probability Distribution - Additional Example → EXAMPLE 2 Suppose that the weekly average number of accidents on a stretch of motorway is 5. (a) What is the probability there will be 2 accidents next week? (b) What is the probability there will be more than 4 accidents next week? (a) We want P(X = 2) where X = “the number of accidents next week”. Using the EXCEL POISSON function, we simply type

Dr. Erwin George, Dept. of Mathematics

STAT1029

Poisson Probability Distribution - Additional Example → EXAMPLE 2 Suppose that the weekly average number of accidents on a stretch of motorway is 5. (a) What is the probability there will be 2 accidents next week? (b) What is the probability there will be more than 4 accidents next week? (a) We want P(X = 2) where X = “the number of accidents next week”. Using the EXCEL POISSON function, we simply type = POISSON(2, 5, FALSE ) and the answer comes out as

Dr. Erwin George, Dept. of Mathematics

STAT1029

Poisson Probability Distribution - Additional Example → EXAMPLE 2 Suppose that the weekly average number of accidents on a stretch of motorway is 5. (a) What is the probability there will be 2 accidents next week? (b) What is the probability there will be more than 4 accidents next week? (a) We want P(X = 2) where X = “the number of accidents next week”. Using the EXCEL POISSON function, we simply type = POISSON(2, 5, FALSE ) and the answer comes out as 0.084224 .

Dr. Erwin George, Dept. of Mathematics

STAT1029

Poisson Probability Distribution - Additional Example → EXAMPLE 2 Suppose that the weekly average number of accidents on a stretch of motorway is 5. (a) What is the probability there will be 2 accidents next week? (b) What is the probability there will be more than 4 accidents next week? (a) We want P(X = 2) where X = “the number of accidents next week”. Using the EXCEL POISSON function, we simply type = POISSON(2, 5, FALSE ) and the answer comes out as 0.084224 . (b) The EXCEL POISSON function only gives P(X = a) or P(x ≤ b) for numbers a and b (it does not give P(X > c) or P(X ≥ d) for numbers c and d). So instead of calculating P(X > 4) directly, we consider the NEGATION OF X > 4 which is

Dr. Erwin George, Dept. of Mathematics

STAT1029

Poisson Probability Distribution - Additional Example → EXAMPLE 2 Suppose that the weekly average number of accidents on a stretch of motorway is 5. (a) What is the probability there will be 2 accidents next week? (b) What is the probability there will be more than 4 accidents next week? (a) We want P(X = 2) where X = “the number of accidents next week”. Using the EXCEL POISSON function, we simply type = POISSON(2, 5, FALSE ) and the answer comes out as 0.084224 . (b) The EXCEL POISSON function only gives P(X = a) or P(x ≤ b) for numbers a and b (it does not give P(X > c) or P(X ≥ d) for numbers c and d). So instead of calculating P(X > 4) directly, we consider the NEGATION OF X > 4 which is X ≤ 4.

Dr. Erwin George, Dept. of Mathematics

STAT1029

Poisson Probability Distribution - Additional Example → EXAMPLE 2 Suppose that the weekly average number of accidents on a stretch of motorway is 5. (a) What is the probability there will be 2 accidents next week? (b) What is the probability there will be more than 4 accidents next week? (a) We want P(X = 2) where X = “the number of accidents next week”. Using the EXCEL POISSON function, we simply type = POISSON(2, 5, FALSE ) and the answer comes out as 0.084224 . (b) The EXCEL POISSON function only gives P(X = a) or P(x ≤ b) for numbers a and b (it does not give P(X > c) or P(X ≥ d) for numbers c and d). So instead of calculating P(X > 4) directly, we consider the NEGATION OF X > 4 which is X ≤ 4. And P(X > 4) = 1 − P(X ≤ 4) =

Dr. Erwin George, Dept. of Mathematics

STAT1029

Poisson Probability Distribution - Additional Example → EXAMPLE 2 Suppose that the weekly average number of accidents on a stretch of motorway is 5. (a) What is the probability there will be 2 accidents next week? (b) What is the probability there will be more than 4 accidents next week? (a) We want P(X = 2) where X = “the number of accidents next week”. Using the EXCEL POISSON function, we simply type = POISSON(2, 5, FALSE ) and the answer comes out as 0.084224 . (b) The EXCEL POISSON function only gives P(X = a) or P(x ≤ b) for numbers a and b (it does not give P(X > c) or P(X ≥ d) for numbers c and d). So instead of calculating P(X > 4) directly, we consider the NEGATION OF X > 4 which is X ≤ 4. And P(X > 4) = 1 − P(X ≤ 4) = 1 − POISSON(4, 5, TRUE ) =
Dr. Erwin George, Dept. of Mathematics STAT1029

Poisson Probability Distribution - Additional Example → EXAMPLE 2 Suppose that the weekly average number of accidents on a stretch of motorway is 5. (a) What is the probability there will be 2 accidents next week? (b) What is the probability there will be more than 4 accidents next week? (a) We want P(X = 2) where X = “the number of accidents next week”. Using the EXCEL POISSON function, we simply type = POISSON(2, 5, FALSE ) and the answer comes out as 0.084224 . (b) The EXCEL POISSON function only gives P(X = a) or P(x ≤ b) for numbers a and b (it does not give P(X > c) or P(X ≥ d) for numbers c and d). So instead of calculating P(X > 4) directly, we consider the NEGATION OF X > 4 which is X ≤ 4. And P(X > 4) = 1 − P(X ≤ 4) = 1 − POISSON(4, 5, TRUE ) = 0.440493 .
Dr. Erwin George, Dept. of Mathematics STAT1029

Binomial Probability Distribution - Additional Example

→ EXAMPLE 3 On average, 55% of marriages end in divorce. What is the probability that in a room of 100 married couples, more than 50 will stay married.

Dr. Erwin George, Dept. of Mathematics

STAT1029

Binomial Probability Distribution - Additional Example

→ EXAMPLE 3 On average, 55% of marriages end in divorce. What is the probability that in a room of 100 married couples, more than 50 will stay married.
ANSWER Here we may assume the likelihood of staying married follows a binomial distribution. The “experiment” is getting married, and it makes sense to define “success” as staying married while “failure” is getting divorced.

Dr. Erwin George, Dept. of Mathematics

STAT1029

Binomial Probability Distribution - Additional Example

→ EXAMPLE 3 On average, 55% of marriages end in divorce. What is the probability that in a room of 100 married couples, more than 50 will stay married.
ANSWER Here we may assume the likelihood of staying married follows a binomial distribution. The “experiment” is getting married, and it makes sense to define “success” as staying married while “failure” is getting divorced. So the probability of success is

Dr. Erwin George, Dept. of Mathematics

STAT1029

Binomial Probability Distribution - Additional Example

→ EXAMPLE 3 On average, 55% of marriages end in divorce. What is the probability that in a room of 100 married couples, more than 50 will stay married.
ANSWER Here we may assume the likelihood of staying married follows a binomial distribution. The “experiment” is getting married, and it makes sense to define “success” as staying married while “failure” is getting divorced. So the probability of success is 0.45.

Dr. Erwin George, Dept. of Mathematics

STAT1029

Binomial Probability Distribution - Additional Example

→ EXAMPLE 3 On average, 55% of marriages end in divorce. What is the probability that in a room of 100 married couples, more than 50 will stay married.
ANSWER Here we may assume the likelihood of staying married follows a binomial distribution. The “experiment” is getting married, and it makes sense to define “success” as staying married while “failure” is getting divorced. So the probability of success is 0.45. The other information we need to know in order to use the EXCEL BINOMDIST function is the number of trials (= 100 marriages in the room) and the value of interest (=50).

Dr. Erwin George, Dept. of Mathematics

STAT1029

Binomial Probability Distribution - Additional Example

→ EXAMPLE 3 On average, 55% of marriages end in divorce. What is the probability that in a room of 100 married couples, more than 50 will stay married.
ANSWER Here we may assume the likelihood of staying married follows a binomial distribution. The “experiment” is getting married, and it makes sense to define “success” as staying married while “failure” is getting divorced. So the probability of success is 0.45. The other information we need to know in order to use the EXCEL BINOMDIST function is the number of trials (= 100 marriages in the room) and the value of interest (=50). So we want P(X > 50), where X = number of couples which stay married.

Dr. Erwin George, Dept. of Mathematics

STAT1029

Binomial Probability Distribution - Additional Example

→ EXAMPLE 3 On average, 55% of marriages end in divorce. What is the probability that in a room of 100 married couples, more than 50 will stay married.
ANSWER Here we may assume the likelihood of staying married follows a binomial distribution. The “experiment” is getting married, and it makes sense to define “success” as staying married while “failure” is getting divorced. So the probability of success is 0.45. The other information we need to know in order to use the EXCEL BINOMDIST function is the number of trials (= 100 marriages in the room) and the value of interest (=50). So we want P(X > 50), where X = number of couples which stay married. Just as for the POISSON function in EXCEL, BINOMDIST only gives P(X = a) or P(X ≤ b) for constants a and b. So instead of calculation P(X > 50) directly, we consider the NEGATION of X > 50 which is

Dr. Erwin George, Dept. of Mathematics

STAT1029

Binomial Probability Distribution - Additional Example

→ EXAMPLE 3 On average, 55% of marriages end in divorce. What is the probability that in a room of 100 married couples, more than 50 will stay married.
ANSWER Here we may assume the likelihood of staying married follows a binomial distribution. The “experiment” is getting married, and it makes sense to define “success” as staying married while “failure” is getting divorced. So the probability of success is 0.45. The other information we need to know in order to use the EXCEL BINOMDIST function is the number of trials (= 100 marriages in the room) and the value of interest (=50). So we want P(X > 50), where X = number of couples which stay married. Just as for the POISSON function in EXCEL, BINOMDIST only gives P(X = a) or P(X ≤ b) for constants a and b. So instead of calculation P(X > 50) directly, we consider the NEGATION of X > 50 which is X ≤ 50.

Dr. Erwin George, Dept. of Mathematics

STAT1029

Binomial Probability Distribution - Additional Example

→ EXAMPLE 3 On average, 55% of marriages end in divorce. What is the probability that in a room of 100 married couples, more than 50 will stay married.
ANSWER Here we may assume the likelihood of staying married follows a binomial distribution. The “experiment” is getting married, and it makes sense to define “success” as staying married while “failure” is getting divorced. So the probability of success is 0.45. The other information we need to know in order to use the EXCEL BINOMDIST function is the number of trials (= 100 marriages in the room) and the value of interest (=50). So we want P(X > 50), where X = number of couples which stay married. Just as for the POISSON function in EXCEL, BINOMDIST only gives P(X = a) or P(X ≤ b) for constants a and b. So instead of calculation P(X > 50) directly, we consider the NEGATION of X > 50 which is X ≤ 50. And P(X > 50) = 1 − P(X ≤ 50) =

Dr. Erwin George, Dept. of Mathematics

STAT1029

Binomial Probability Distribution - Additional Example

→ EXAMPLE 3 On average, 55% of marriages end in divorce. What is the probability that in a room of 100 married couples, more than 50 will stay married.
ANSWER Here we may assume the likelihood of staying married follows a binomial distribution. The “experiment” is getting married, and it makes sense to define “success” as staying married while “failure” is getting divorced. So the probability of success is 0.45. The other information we need to know in order to use the EXCEL BINOMDIST function is the number of trials (= 100 marriages in the room) and the value of interest (=50). So we want P(X > 50), where X = number of couples which stay married. Just as for the POISSON function in EXCEL, BINOMDIST only gives P(X = a) or P(X ≤ b) for constants a and b. So instead of calculation P(X > 50) directly, we consider the NEGATION of X > 50 which is X ≤ 50. And P(X > 50) = 1 − P(X ≤ 50) =

1 − BINOMDIST(50, 100, 0.45, TRUE ) =

Dr. Erwin George, Dept. of Mathematics

STAT1029

Binomial Probability Distribution - Additional Example

→ EXAMPLE 3 On average, 55% of marriages end in divorce. What is the probability that in a room of 100 married couples, more than 50 will stay married.
ANSWER Here we may assume the likelihood of staying married follows a binomial distribution. The “experiment” is getting married, and it makes sense to define “success” as staying married while “failure” is getting divorced. So the probability of success is 0.45. The other information we need to know in order to use the EXCEL BINOMDIST function is the number of trials (= 100 marriages in the room) and the value of interest (=50). So we want P(X > 50), where X = number of couples which stay married. Just as for the POISSON function in EXCEL, BINOMDIST only gives P(X = a) or P(X ≤ b) for constants a and b. So instead of calculation P(X > 50) directly, we consider the NEGATION of X > 50 which is X ≤ 50. And P(X > 50) = 1 − P(X ≤ 50) =

1 − BINOMDIST(50, 100, 0.45, TRUE ) = 0.182728 .

Dr. Erwin George, Dept. of Mathematics

STAT1029