Question

Mrs. Jones buys two toys for her son. The probability that the first toy is defective is 1/3, and the probability that the second toy is defective given that the first toy is defective is 1/5. What is the probability that both toys are defective?

Answer 1

Answer: $\frac{1}{15}$

Step-by-step explanation:

Let F denote the event of first  toy is defective and S denote the second toy is defective .

Given: The probability that the first toy is defective $P(F)=\frac{1}{3}$

The probability that the second toy is defective given that the first toy is defective  $P(S|F)=\frac{1}{5}$

The formula to calculate the conditional probability is given by :-

$P(S|F)=\frac{P(S\cap F)}{P(F)}\\\\\Rightarrow P(S\cap F)=P(S|F)\times P(F)\\\\\Rightarrow\ P(S\cap F)=\frac{1}{3}\times \frac{1}{5}=\frac{1}{15}$

Hence, the probability that both toys are defective=$\frac{1}{15}$

Answer 2

Answer:

Probability=$\frac{1}{15}$

Step-by-step explanation:

As it is given that

Probability of toy A is defective is =P(A) = $\frac{1}{3}$

Probability of toy b is defective if A is defective = P (B)=$\frac{1}{5}$

WE have to find the P(A n B)

By the law of Probability

P(A n B) = P (A).P(B)

putting the values given to us

P(AnB)=$\frac{1}{3}$ * $\frac{1}{5}$

Probability=$\frac{1}{15}$