Question

Joe 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 eggs 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

The correct option is (b).

Step-by-step solution

Given Informaiton

The number of eggs carrying germs is one out of four.

His egg consumption is less than three.

Explanation for correct option

The probability of success is:

$P\left(\text{Success}\right)=\frac{1}{4}$

Number of eggs $=3$

Since, the probabilities and number of eggs are independent of each other. Thus, binomial distribution is being used with the parameters $p=\frac{1}{4}$and $n=3$.

Hence, the correct option is (b).

Explanation for incorrect option

(a) The number of contaminated eggs when Joe uses 3 eggs chosen at random has the following distribution will not be binomial$n=4andp=1/4$

(c) The number of contaminated eggs when Joe uses 3 eggs chosen at random has the following distribution will not be binomial 3

(d) The number of contaminated eggs when Joe uses 3 eggs chosen at random has the following distribution will not be geometric$p=1/4$

(e) The number of contaminated eggs when Joe uses 3 eggs chosen at random has the following distribution will not be geometric$p=1/3$