Question

Let $A,B,C$ be three events. If the probability of occurring exactly one event out of $A$ and $B$ is $1-a$, out of $B$ and $C$ and $A$ is $1-2a$ and that of occurring three events simultaneously is $a^2$, then the probability that at least one out of $A,B,C$ will occur is

• A. $\frac{1}{2}$
• B. Greater than $\frac{1}{2}$
• C. Less than $\frac{1}{2}$
• D. Greater than $\frac{3}{4}$

Answer 1

The correct option is B

Greater than $\frac{1}{2}$

Explanation for the correct option:

Finding the probability that at least one out of $A,B,C$ will occur:

Given that,

Probability of getting exactly one event out of $A$ and $B$ $=1-a$

Therefore,

$$\begin{gathered} \Rightarrow P\left[\right(A\cap B’)\cup (A’\cap B\left)\right]=1-a \\ \Rightarrow P(A\cup B)–P(A\cap B)=1-a \\ \Rightarrow \left(P\left(A\right)+P\left(B\right)–P(A\cap B)\right)–P(A\cap B)=1-a \\ \Rightarrow P\left(A\right)+P\left(B\right)–2P(A\cap B)=1-a...\left(i\right) \end{gathered}$$

Similarly,

Probability of getting exactly one event out of $B$ and $C$ $=1-2a$

$P\left(B\right)+P\left(C\right)–2P(B\cap C)=1-2a...\left(ii\right)$

$P\left(C\right)+P\left(A\right)–2P(C\cap A)=1-a..\left(iii\right)$

And given that that of occurring three events simultaneously is $a^2$,

$P(A\cap B\cap C)=a^2..\left(iv\right)$

As we know,

$$\begin{gathered} P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)–P(A\cup B)–P(B\cup C)-P(C\cup A)+P(A\cap B\cap C) \\ =\frac{1}{2}\left[P\right(A)+P(B)–2P(B\cap C)+P(B)+P(C)–2P(B\cap C)+P(C)+P(A)–2P(C\cap A\left)\right]+P(A\cap B\cap C) \\ =\frac{1}{2}\left[1-a+1-2a+1-a\right]+a^2 \end{gathered}$$

Now from $\left(i\right),\left(ii\right),\left(iii\right)\&\left(iv\right)$

$$\begin{gathered} =\frac{3}{2}–2a+a^2 \\ =\frac{1}{2}+{(a-1)}^2>\frac{1}{2} \end{gathered}$$

Hence, the correct is $option\left(B\right)$