In a region, there is a 0.7 probability chance that a randomly selected person o

Question

In a region, there is a 0.7 probability chance that a randomly selected person of the population has brown eyes. Assume 15 people are randomly selected. Complete parts (a) through (d) below. a. Find the probability that all of the selected people have brown eyes. The probability that all of the 15 selected people have brown eyes is 0.005 (Round to three decimal places as needed.) b. Find the probability that exactly 14 of the selected people have brown eyes. The probability that exactly 14 of the selected people have brown eyes is 0.031 (Round to three decimal places as needed.) c. Find the probability that the number of selected people that have brown eyes is 13 or more. The probability that the number of selected people that have brown eyes is 13 or more is Round to three decimal places as needed.)

Explanation / Answer

Question 1

Solution:

Here, we have to use binomial distribution.

Formula for binomial probability is given as below:

P(X=x) = nCx*p^x*q^(n – x)

We are given

n = 15

p = 0.7

q = 1 – p = 1 – 0.7 = 0.3

Part a

We have to find P(X=15)

P(X=15) = 15C15*0.7^15*0.3^0

P(X=15) = 0.004748

Required probability = 0.005

Part b

We have to find P(X=14)

P(X=14) = 15C14*0.7^14*0.3^1

P(X=14) = 0.03052

Required probability = 0.031

Part c

Here, we have to find P(X13)

P(X13) = P(X=13) + P(X=14) + P(X=15)

P(X=13) = 15C13*0.7^13*0.3^2

P(X=13) = 0.09156

P(X=14) = 15C14*0.7^14*0.3^1

P(X=14) = 0.03052

P(X=15) = 15C15*0.7^15*0.3^0

P(X=15) = 0.004748

P(X13) = P(X=13) + P(X=14) + P(X=15)

P(X13) = 0.09156 + 0.03052 + 0.004748

P(X13) = 0.126828

Required probability = 0.127

Question 2

The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: Game winning probability is 50%.

Alternative hypothesis: Ha: Game winning probability is not 50%.

In symbolic form,

H0: p = 0.5 versus Ha: p 0.5

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