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MHTCET 15 Apr 2026 Shift 1 Paper

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Question : 150 of 150

Marks: +1, -0

If A and B are two events such that

P(A)=\frac{2}{3}, P(B)=\frac{1}{2}

P(A \mid B)=\frac{2}{3}

P\left(A^{\prime} \cup B\right)+P\left(A \cup B^{\prime}\right)=

$\frac{2}{3}$
1
$\frac{3}{2}$
$\frac{5}{6}$

Solution:

Concept:

Use conditional probability to find

P(A \cap B)

, then apply complement and union laws.

Explanation:

Given

P(A)=\frac{2}{3}

,

P(B)=\frac{1}{2}

, and

P(A \mid B)=\frac{2}{3}

.

By the conditional probability formula:

P(A \mid B)=\frac{P(A \cap B)}{P(B)}

So,

\frac{2}{3}=\frac{P(A \cap B)}{\frac{1}{2}}

P(A \cap B)=\frac{2}{3}\times \frac{1}{2}=\frac{1}{3}

Now,

P\left(A \cap B^{\prime}\right)=P(A)-P(A \cap B)

P\left(A \cap B^{\prime}\right)=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}

Using the complement rule:

P\left(A^{\prime} \cup B\right)=1-P\left(A \cap B^{\prime}\right)

P\left(A^{\prime} \cup B\right)=1-\frac{1}{3}=\frac{2}{3}

Similarly,

P\left(A^{\prime} \cap B\right)=P(B)-P(A \cap B)

P\left(A^{\prime} \cap B\right)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}

So,

P\left(A \cup B^{\prime}\right)=1-P\left(A^{\prime} \cap B\right)

P\left(A \cup B^{\prime}\right)=1-\frac{1}{6}=\frac{5}{6}

Therefore,

P\left(A^{\prime} \cup B\right)+P\left(A \cup B^{\prime}\right)=\frac{2}{3}+\frac{5}{6}

=\frac{4}{6}+\frac{5}{6}=\frac{9}{6}=\frac{3}{2}

Answer:

\frac{3}{2}

Hence, the correct option is C.

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