Question

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; $n=4$and$p=1/4$

(b) binomial; $n=3$and$p=1/4$

(c) binomial; $n=3$and$p=1/3$

(d) geometric;$p=1/4$

(e) geometric;$p=1/3$

Short answer

(b) Binomial distribution with $n=3$ and$p=1/4$

Step-by-step solution

Given Information 

Eggs contains salmonella bacteria$=1outof4$

Eggs not used for cooking$=morethan3$

Explanation 

An assumption is made that probability is $1$out of $4$

Randomly, $3$are chosen

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

$P\left(\text{win}\right)=\frac{\mathrm{\#}\text{of favorable outcomes}}{\mathrm{\#}\text{of possible outcomes}}=\frac{1}{4}$

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

$n=3$