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;$p=1/3$

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

His egg consumption is less than three.

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).

(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$p=1/3$