Question

Two events $A$ and$B$ are such that:$P\left(A\right)=\frac{8}{15},P\left(B\right)=\frac{1}{3},P\left(A|B\right)=\frac{1}{5}$

Calculate the probability that:

$a)$Both events occur

$b)$Only one of the two events occurs.

$c)$Neither event occurs

Answer 1

Step-1: Finding the probability that both events occur:

The probability that both events occur is $P\left(A\cap B\right)$

As we know $P\left(A\cap B\right)=P\left(A|B\right)·P\left(B\right)$

$P\left(A\cap B\right)=\frac{1}{5}·\frac{1}{3}=\frac{1}{15}$

Therefore, the probability that both events occur is $\frac{1}{15}$.

Step-2: Finding the probability that only one of the two events occurs

The probability that only one of the two events occurs is $P\left(A\cup B\right)-P\left(A\cap B\right)$

As we know $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$$\begin{gathered} P\left(A\cup B\right)=\frac{8}{15}+\frac{1}{3}-\frac{1}{15} \\ =\frac{8+5-1}{15} \\ =\frac{12}{15} \end{gathered}$$

Now,

$$\begin{gathered} P\left(A\cup B\right)-P\left(A\cap B\right) \\ =\frac{12}{15}-\frac{1}{15} \\ =\frac{11}{15} \end{gathered}$$

Therefore, the probability that only one of the two events occurs is $\frac{11}{15}$.

Step-3: Finding the probability that neither event occurs

The probability that neither event occurs is $1-P\left(A\cup B\right)$

So,

$$\begin{gathered} 1-P\left(A\cup B\right) \\ =1-\frac{12}{15} \\ =\frac{15-12}{15} \\ =\frac{3}{15} \\ =\frac{1}{5} \end{gathered}$$

Therefore, the probability that neither event occurs is$\frac{1}{5}$.