Question

flipping a coin 5 times, with 95 percent chance of being head and 5 percent chance of being tail and you win 20 dollars for each head, and loses 20 dollars for each tail. 1)What is the probability of getting one head? 2) two heads? 3) three heads?4) four heads? 5) five heads? 6)What is the probability of getting at least one head? 7) at least two heads?8) at least three heads? 9) at least four heads? 10) find the expected value for question 1,2 and 3.

Answer 1

p = probability of getting head = 0.95

1 - p = probability of getting tail = 1 - 0.95 = 0.05

n = number of trials = 5

Here n is fixed and the trials are independent.

Also probability of getting head or tail is constant for each trials.

So we can use binomial distribution to find the probabilities.

Let X = number of heads in the 5 trials.

Let's use excel:

The formulae used on the above excel sheet are as follows:

6) What is the probability of getting at least one head?

P( X >= 1) = 1 - P( X = 0) = "=1-BINOMDIST(0,5,0.95,0)" = 0.999999688

7) at least two heads?

P( X >= 2) = 1 - P( X <= 1) = "=1-BINOMDIST(1,5,0.95,1)" = 0.99997

8) at least three heads?

P( X >= 3) = 1 - P( X <= 2) =  "=1-BINOMDIST(2,5,0.95,1)" = 0.998842

9) at least four heads?

P( X >= 4) = 1 - P( X <= 3) =  "=1-BINOMDIST(3,5,0.95,1)" = 0.773781

10 ) Expected value of question (1)

In this question we get one head and 4 tails

So net gain is 20 - ( 4* 20) = -60

So expected value = -60 * P( X = 1) = -60*0.000029688 = -0.00178125

Expected value of question (2)

In this question we get two head and 3 tails

So net gain is 2*20 - ( 3* 20) = -20

So expected value = -20 * P( X = 2) = 20*0.001128125 = -0.0225625

Expected value of question (3)

In this question we get 3 head and 2 tails

So net gain is 3*20 - ( 2* 20) = 20

So expected value = 20 * P( X = 3) = 20*021434375 = 0.4286875