James reads that 1 out of 6 eggs contains salmonella bacteria. So he never uses more than 5 eggs in cooking. If eggs do or don't contain salmonella independently of each other, the number of contaminated eggs when James uses 5 chosen at random has the distributionA. binomial with n = 6 and p = 1/6. B. binomial with n = 5 and p = 1/6. C. binomial with n = 5 and p = 1/5.

Question

Carl Spence

Mathematics

14 April, 17:26

James reads that 1 out of 6 eggs contains salmonella bacteria. So he never uses more than 5 eggs in cooking. If eggs do or don't contain salmonella independently of each other, the number of contaminated eggs when James uses 5 chosen at random has the distributionA. binomial with n = 6 and p = 1/6. B. binomial with n = 5 and p = 1/6. C. binomial with n = 5 and p = 1/5.

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Rachael · Answer 1

B. binomial with n = 5 and p = 1/6

Step-by-step explanation:

Given that James reads that 1 out of 6 eggs contains salmonella bacteria.

So he never uses more than 5 eggs in cooking.

If eggs do or don't contain salmonella independently of each other, then

probability for any one egg to contain salmonella bacteria is the constant 1/6.

Also there are only two outcomes, either bacteria present or not.

Hence the number of contaminated eggs when James uses 5 chosen at random has the distribution is Binomial

Here n = no of trials = no of eggs he uses = 5

Probability = 1/6

So option B is right.