Answer:
[tex]P(A\textrm{ and }B)=\frac{3}{14}[/tex]
Step-by-step explanation:
Given:
[tex]P(A)=\frac{4}{7}[/tex]
[tex]P(B|A)=\frac{3}{8}[/tex]
We know that, conditional probability of B given that A has occurred is given as:
[tex]P(B|A)=\frac{P(A\cap B}{P(A)}[/tex]. Expressing this in terms of [tex]P(A\cap B)[/tex], we get
[tex]P(A\cap B)=P(B|A)\times P(A)[/tex]
Plug in the known values and solve for [tex]P(A\cap B)[/tex]. This gives,
[tex]P(A\cap B)=P(B|A)\times P(A)\\P(A\cap B)=\frac{3}{8}\times \frac{4}{7}\\P(A\cap B)=\frac{12}{56}=\frac{3}{14}[/tex]
Therefore, the probability of events A and B occurring is [tex]\frac{3}{14}[/tex].