Answer:
Probability=[tex]\frac{1}{15}[/tex]
Step-by-step explanation:
As it is given that
Probability of toy A is defective is =P(A) = [tex]\frac{1}{3}[/tex]
Probability of toy b is defective if A is defective = P (B)=[tex]\frac{1}{5}[/tex]
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)=[tex]\frac{1}{3}[/tex] * [tex]\frac{1}{5}[/tex]
Probability=[tex]\frac{1}{15}[/tex]
Answer: [tex]\frac{1}{15}[/tex]
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 [tex]P(F)=\frac{1}{3}[/tex]
The probability that the second toy is defective given that the first toy is defective [tex]P(S|F)=\frac{1}{5}[/tex]
The formula to calculate the conditional probability is given by :-
[tex]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}[/tex]
Hence, the probability that both toys are defective=[tex]\frac{1}{15}[/tex]
