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101. Job reads that 1 out of 4 eggs contains salmonella bacteria. So he never uses more than 3 eggs in cooking. If eggs do or don't contain salmonella independently of each other, the number of contaminated eggs when Joe uses 3 chosen at random has the following distribution:

(a) binomial; n=4andp=1/4

(b) binomial; n=3andp=1/4

(c) binomial; n=3andp=1/3

(d) geometric;p=1/4

(e) geometric;p=1/3

Short Answer

Expert verified

(b) Binomial distribution with n=3 andp=1/4

Step by step solution

01

Given Information 

Eggs contains salmonella bacteria=1outof4

Eggs not used for cooking=morethan3

02

Explanation 

An assumption is made that probability is 1out of 4

Randomly, 3are chosen

By dividing the number of favorable outcomes by the possible outcomes, we arrive at probability:

P(win)=#of favorable outcomes#of possible outcomes=14

3are chosen at random, So the sample size is:

n=3

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