In the Redskins stadium 40% are Dallas cowboy’s fans in 6 tries the camera shows fans. What is the probability that a Dallas fan will be seen on television?

n=6 p=.40 q=.60
My variables are 0,1,2,3,4,5,6
P(x) = nCx(p ^x)*(q^ n-x) nCx=[( n! / (n-x)! ) / x! ] = 6! / (6-1)! *1! = 720/120 = 6 p ^ x = .40 ^ 1=.40 q ^ n – x = .60 ^ 6 – 1 = .60 ^ 5 = .07776
6 * .40* .0776 = .186 chance of one of the fans being a Dallas cowboy fan. Oh Yeah…

| Example: Toss a coin for 12 times. What is the probability of getting exactly 7 heads.

Step 1: Here,
Number of trials n = 12
Number of success r = 7 (since we define getting a head as success)
Probability of success on any single trial p = 0.5

Step 2: To Calculate nCr formula is used. nCr = ( n! / (n-r)! ) / r!
= ( 12! / (12-7)! ) / 7!
= ( 12! / 5! ) / 7!
= ( 479001600 / 120 ) / 5040
= ( 3991680 / 5040 )
= 792

Step 3: Find pr. pr = 0.57
= 0.0078125

Step 4: To Find (1-p)n-r Calculate 1-p and n-r.
1-p = 1-0.5 = 0.5 n-r = 12-7 = 5

Step 5: Find (1-p)n-r.
= 0.55 = 0.03125

Step 6: Solve P(X = r) = nCr p r (1-p)n-r
= 792 × 0.0078125 × 0.03125
= 0.193359375

The probability of getting exactly 7 heads is 0.19
I am keeping my fingers crossed that this is right and correct.
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