If A and B are two events such that P(A) = 0.7, P(B) = 0.4 and P ( A B) = 0.5, where B denotes the complement of B, then P ( B ( A B)) is equal to

Given: P(A) = 0.7, P(B) = 0.4 and P(A B) = 0.5. Find: P (B (A B)). Use conditional probability: P (B (A B)) = P (B (A B))P(A B) First, find P(A B) using P(A B) = P(A) - P(A B) So, 0.5 = 0.7 - P(A B) P(A B) = 0.2 Now, P( B) = 1 - P(B) = 1 - 0.4 = 0.6 Then, P(A B) = P(A) + P( B) - P(A B) = 0.7 + 0.6 - 0.5 = 0.8 Also, B (A B) = (B A) (B B) Since B B =, P (B (A B)) = P(A B) = 0.2 Therefore, P (B (A B)) = 0.20.8 = 14 Therefore, the correct option is B. Given: P(A) = 0.7, P(B) = 0.4 and P(A B) = 0.5. Find: P (B (A B)). From P(A B) = P(A) - P(A B) we get 0.5 = 0.7 - P(A B) P(A B) = 0.2 Next, P(A B) = P(A) + P( B) - P(A B) with P( B) = 1 - 0.4 = 0.6 Hence, P(A B) = 0.7 + 0.6 - 0.5 = 0.8 Now evaluate the numerator: P(B (A B)) = P((B A) (B B)) = P(A B) + P() = 0.2 + 0 = 0.2 Finally, P (B (A B)) = P(B (A B))P(A B) = 0.20.8 = 14 Thus, the required probability is 14, so the correct option is B.

If A and B are two events such that P(A) = 0.7, P(B) = 0.4 and P… — Solution | JEE Mathematics

MCQMediumJEE 2025Conditional Probability & Bayes Theorem

If $A$ and $B$ are two events such that $P(A) = 0.7$ , $P(B) = 0.4$ and $P\left( A \cap \overline{B} \right) = 0.5$ , where $\overline{B}$ denotes the complement of $B$ , then $P\left( B \mid \left( A \cup \overline{B} \right) \right)$ is equal to

A
$\frac{1}{2}$
B
$\frac{1}{4}$
C
$\frac{1}{3}$
D
$\frac{1}{6}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $P(A) = 0.7$ , $P(B) = 0.4$ and $P(A \cap \overline{B}) = 0.5$ .

Find: $P\left(B \mid (A \cup \overline{B})\right)$ .

Use conditional probability:

P\left(B \mid (A \cup \overline{B})\right) = \frac{P\left(B \cap (A \cup \overline{B})\right)}{P(A \cup \overline{B})}

First, find $P(A \cap B)$ using

P(A \cap \overline{B}) = P(A) - P(A \cap B)

So,

0.5 = 0.7 - P(A \cap B)

P(A \cap B) = 0.2

Now,

P(\overline{B}) = 1 - P(B) = 1 - 0.4 = 0.6

Then,

P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B})

= 0.7 + 0.6 - 0.5 = 0.8

Also,

B \cap (A \cup \overline{B}) = (B \cap A) \cup (B \cap \overline{B})

Since $B \cap \overline{B} = \emptyset$ ,

P\left(B \cap (A \cup \overline{B})\right) = P(A \cap B) = 0.2

Therefore,

P\left(B \mid (A \cup \overline{B})\right) = \frac{0.2}{0.8} = \frac{1}{4}

Therefore, the correct option is B.

Stepwise Set-Based Evaluation

Given: $P(A) = 0.7$ , $P(B) = 0.4$ and $P(A \cap \overline{B}) = 0.5$ .

Find: $P\left(B \mid (A \cup \overline{B})\right)$ .

From

P(A \cap \overline{B}) = P(A) - P(A \cap B)

we get

0.5 = 0.7 - P(A \cap B)

P(A \cap B) = 0.2

Next,

P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B})

with

P(\overline{B}) = 1 - 0.4 = 0.6

Hence,

P(A \cup \overline{B}) = 0.7 + 0.6 - 0.5 = 0.8

Now evaluate the numerator:

P(B \cap (A \cup \overline{B})) = P((B \cap A) \cup (B \cap \overline{B}))

= P(A \cap B) + P(\emptyset) = 0.2 + 0 = 0.2

Finally,

P\left(B \mid (A \cup \overline{B})\right) = \frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})} = \frac{0.2}{0.8} = \frac{1}{4}

Thus, the required probability is $\frac{1}{4}$ , so the correct option is B.

Common mistakes

Using $P(A \cap \overline{B})$ as $P(A \cap B)$ is incorrect because these are different regions of the Venn diagram. Instead, use $P(A \cap \overline{B}) = P(A) - P(A \cap B)$ to find $P(A \cap B)$ first.
Writing the denominator as $P(A \cup B)$ instead of $P(A \cup \overline{B})$ changes the event completely. The conditioning event must be copied exactly before applying the conditional probability formula.
Assuming $B \cap (A \cup \overline{B}) = B$ is wrong because only the part of $B$ lying inside $A \cup \overline{B}$ is counted. Use set distribution: $B \cap (A \cup \overline{B}) = (B \cap A) \cup (B \cap \overline{B})$ , and then note that $B \cap \overline{B} = \emptyset$ .

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Answer

Step-by-step solution

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