Question

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

(a) $0.84$.

(b) $0.68$.

(c) $0.58$.

(d) $0.42$.

(e) $0.30$.

Short answer

(c) $P$(at least$1$ salmonella)$\approx 0.58$

Step-by-step solution

Given Information 

Joe's Eggs contain salmonella bacteria$1outof3$

Eggs not used for cooking $=morethan3$

Explanation 

Result exercise $101$:

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

Complement rule:

$P(notA)=1-P\left(A\right)$

Find the probability of an egg not containing salmonella:

$P\left(\text{not salmonella}\right)=1-P\left(\text{salmonella}\right)=1-\frac{1}{4}=\frac{3}{4}$

Apply the Multiplication rule (if $A$and $B$are independent):

$P\left(A\text{and}B\right)=P\left(A\right)\times P\left(B\right)$

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

$P\left(3\text{not salmonella}\right)=P(\text{not salmonella}{)}^3$

$={\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)}^3$

$=\frac{{\displaystyle 27}}{{\displaystyle 64}}$

$\approx 0.42$

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

$P\left(\text{at least}1\text{salmonella}\right)=1-P\left(3\text{not salmonella}\right)$

$=1-\frac{27}{64}$

$=\frac{37}{64}$

$\approx 0.58$

Hence, the probability is option (c)$0.58$