Question:
Two brothers \(X\) and \(Y\) appeared for an exam. Let \(A\) be the event that \(X\) has passed the exam and \(B\) be the event that \(Y\) has passed. The probability of \(A\) is
\[
\frac17
\]
and of \(B\) is
\[
\frac29.
\]
Then, the probability that both of them pass the exam is
Two brothers \(X\) and \(Y\) appeared for an exam. Let \(A\) be the event that \(X\) has passed the exam and \(B\) be the event that \(Y\) has passed. The probability of \(A\) is
\[
\frac17
\]
and of \(B\) is
\[
\frac29.
\]
Then, the probability that both of them pass the exam is
Show Hint
For independent events \(A\) and \(B\),
\[
P(A\cap B)=P(A)\times P(B).
\]
Use multiplication whenever both events must occur together.
Updated On: Jun 22, 2026
- \(\frac1{63}\)
- \(\frac2{35}\)
- \(\frac2{63}\)
- \(\frac9{14}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Write the given probabilities.
Let \[ A=\text{event that }X\text{ passes} \] and \[ B=\text{event that }Y\text{ passes} \] Given, \[ P(A)=\frac17 \] and \[ P(B)=\frac29 \]
Step 2: Use the multiplication rule for independent events.
Since the performance of one brother does not affect the other, the events are independent.
Therefore, \[ P(A\cap B)=P(A)\times P(B) \]
Step 3: Substitute the values.
\[ P(A\cap B) = \frac17\times\frac29 \] \[ = \frac2{63} \]
Step 4: Final conclusion.
Hence, the probability that both pass the examination is \[ \boxed{\frac2{63}} \]
Let \[ A=\text{event that }X\text{ passes} \] and \[ B=\text{event that }Y\text{ passes} \] Given, \[ P(A)=\frac17 \] and \[ P(B)=\frac29 \]
Step 2: Use the multiplication rule for independent events.
Since the performance of one brother does not affect the other, the events are independent.
Therefore, \[ P(A\cap B)=P(A)\times P(B) \]
Step 3: Substitute the values.
\[ P(A\cap B) = \frac17\times\frac29 \] \[ = \frac2{63} \]
Step 4: Final conclusion.
Hence, the probability that both pass the examination is \[ \boxed{\frac2{63}} \]
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