Question

If events are independents and $P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{3}andP\left(C\right)=\frac{1}{4}$ then $P(A\text{'}\cap B\text{'}\cap C\text{'})$is equal to

Answer 1

Step $1$ : Finding the Probabilities of $P\text{'}\left(A\right),P\text{'}\left(B\right),P\text{'}\left(C\right)$

Given that Events are independent and $P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{3}andP\left(C\right)=\frac{1}{4}$

$$\begin{gathered} P\left(A^{\text{'}}\right)=1-\left(\frac{1}{3}\right)=\frac{2}{3} \\ P\left(B^{\text{'}}\right)=1-\left(\frac{1}{3}\right)=\frac{2}{3} \\ P\left(C^{\text{'}}\right)=1-\left(\frac{1}{4}\right)=\frac{3}{4} \end{gathered}$$

As $A,B,andC$are independent events.

Step $2$ : Finding the required probability

$$\begin{gathered} \Rightarrow P(A^{\text{'}}\cap B^{\text{'}}\cap C^{\text{'}})=P\left(A^{\text{'}}\right).P\left(B^{\text{'}}\right)P\left(C^{\text{'}}\right) \\ \Rightarrow P(A^{\text{'}}\cap B^{\text{'}}\cap C^{\text{'}})=\left(\frac{2}{3}\right).\left(\frac{2}{3}\right).\left(\frac{3}{4}\right) \\ \Rightarrow P(A^{\text{'}}\cap B^{\text{'}}\cap C^{\text{'}})=\frac{12}{36} \\ =\frac{1}{3} \end{gathered}$$

Hence, the value of $\mathit{P}\mathbf{(}\mathit{A}\mathbf{\text{'}}\mathbf{\cap }\mathit{B}\mathbf{\text{'}}\mathbf{\cap }\mathit{C}\mathbf{\text{'}}\mathbf{)}\mathbf{}$ is $\frac{\mathbf{1}}{\mathbf{3}}.$