Question

The probability of happening at least one of the events $A$ and $B$ is $0.6$. If the events $A$and $B$ happens simultaneously with the probability $0.2$, then $P\left(A^{\text{'}}\right)+P\left(B^{\text{'}}\right)=$

• A. $0.4$
• B. $0.8$
• C. $1.4$
• D. $1.2$

Answer 1

The correct option is D

$1.2$

Calculate the required probability

Given, the probability that atleast one of the events $A,B$ happens is $P\left(A\cup B\right)$

So, $P\left(A\cup B\right)=0.6$

Also, the probability of simultaneous happening is $P\left(A\cap B\right)$

then $P\left(A\cap B\right)=0.2$

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

$$\begin{gathered} P\left(A\cup B\right)=P\left(A\right)-P\left(A\cap B\right)+P\left(B\right) \\ \Rightarrow P\left(A\right)+P\left(B\right)=P\left(A\cup B\right)+P\left(A\cap B\right) \\ \Rightarrow P\left(A\right)+P\left(B\right)=0·6+0·2 \\ \Rightarrow P\left(A\right)+P\left(B\right)=0.8 \end{gathered}$$

Now ,$P\left(A\right)+P\left(A^{\text{'}}\right)=1$and $P\left(B\right)+P\left(B^{\text{'}}\right)=1$

$$\begin{gathered} 1-P\left(A^{\text{'}}\right)+1-P\left(B^{\text{'}}\right)=0.8 \\ \Rightarrow 2-P\left(A^{\text{'}}\right)-P\left(B^{\text{'}}\right)=0.8 \\ \Rightarrow P\left(A^{\text{'}}\right)+P\left(B^{\text{'}}\right)=2-0.8 \\ \Rightarrow P\left(A^{\text{'}}\right)+P\left(B^{\text{'}}\right)=1.2 \end{gathered}$$

Therefore, $P\left(A^{\text{'}}\right)+P\left(B^{\text{'}}\right)=1.2$, so option (D) is the correct answer.