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Chapter 6: Q. 102 (page 406)

In the previous exercise, the probability that at least 1of Joe's 3eggs contains salmonella is about

(a) 0.84.

(b) 0.68.

(c) 0.58.

(d) 0.42.

(e) 0.30.

Short Answer

Expert verified

(c) P(at least1 salmonella)鈮�0.58

Step by step solution

01

Given Information 

Joe's Eggs contain salmonella bacteria1outof3

Eggs not used for cooking =morethan3

02

Explanation 

Result exercise 101:

P(salmonella)=p=14

Complement rule:

P(notA)=1-P(A)

Find the probability of an egg not containing salmonella:

P(not salmonella)=1鈭�P(salmonella)=1鈭�14=34

Apply the Multiplication rule (if Aand Bare independent):

P(AandB)=P(A)脳P(B)

It is then likely that all three eggs will be free of salmonella:

P(3not salmonella)=P(not salmonella)3

=343

=2764

鈮�0.42

We can then calculate the probability of finding salmonella in at least one egg using the complement rule:

P(at least1salmonella)=1鈭�P(3not salmonella)

=1鈭�2764

=3764

鈮�0.58

Hence, the probability is option (c)0.58

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