Answer: [tex]\dfrac{1}{15}[/tex]
Step-by-step explanation:
Let A denote the event of first toy is defective and B denote the second toy is defective .
Given: The probability that the first toy is defective [tex]:\ P(F)=\dfrac{1}{3}[/tex]
The probability that the second toy is defective given that the first toy is defective = [tex]P(B|A)=\dfrac{1}{5}[/tex]
The formula to calculate the conditional probability is given by :
[tex]P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow\ P(A\cap B)=P(B|A)\times P(A)\\\\\Rightarrow\ P(A\cap B)=\dfrac{1}{5}\times\dfrac{1}{3}=\dfrac{1}{15}[/tex]
Hence, the required probability : [tex]\dfrac{1}{15}[/tex]
