Question:
If \(A\), \(B\), \(C\) are independent events of a random experiment such that
\[
P(A)=\frac34,\qquad
P(B)=\frac56,\qquad
P(C)=\frac23,
\]
then the probability that exactly one of the events occurs is
If \(A\), \(B\), \(C\) are independent events of a random experiment such that
\[
P(A)=\frac34,\qquad
P(B)=\frac56,\qquad
P(C)=\frac23,
\]
then the probability that exactly one of the events occurs is
Show Hint
For independent events, "exactly one occurs" means:
\[
A B^c C^c
\quad \text{or} \quad
A^c B C^c
\quad \text{or} \quad
A^c B^c C.
\]
Find each probability separately and then add them.
Updated On: Jul 9, 2026
- \(\dfrac14\)
- \(\dfrac{5}{36}\)
- \(\dfrac{5}{12}\)
- \(\dfrac{71}{72}\) \bigskip
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
For independent events, the probability that exactly one event occurs is the sum of the probabilities of the following mutually exclusive cases:
\[
A\cap B^c\cap C^c,
\]
\[
A^c\cap B\cap C^c,
\]
\[
A^c\cap B^c\cap C.
\]
Also,
\[
P(A^c)=1-P(A).
\]
Step 1: Find the probabilities of the complementary events. \[ P(A^c)=1-\frac34=\frac14, \] \[ P(B^c)=1-\frac56=\frac16, \] \[ P(C^c)=1-\frac23=\frac13. \]
Step 2: Find the probability that only \(A\) occurs. Since the events are independent, \[ P(A\cap B^c\cap C^c) = P(A)P(B^c)P(C^c). \] \[ = \frac34\cdot\frac16\cdot\frac13. \] \[ = \frac1{24}. \]
Step 3: Find the probability that only \(B\) occurs. \[ P(A^c\cap B\cap C^c) = P(A^c)P(B)P(C^c). \] \[ = \frac14\cdot\frac56\cdot\frac13. \] \[ = \frac5{72}. \]
Step 4: Find the probability that only \(C\) occurs. \[ P(A^c\cap B^c\cap C) = P(A^c)P(B^c)P(C). \] \[ = \frac14\cdot\frac16\cdot\frac23. \] \[ = \frac1{36}. \]
Step 5: Add the three mutually exclusive cases. \[ P(\text{exactly one event occurs}) = \frac1{24} + \frac5{72} + \frac1{36}. \] Taking LCM \(=72\), \[ = \frac3{72} + \frac5{72} + \frac2{72}. \] \[ = \frac{10}{72}. \] \[ = \frac5{36}. \]
Step 6: Write the final answer. \[ \boxed{\frac5{36}} \]
Step 1: Find the probabilities of the complementary events. \[ P(A^c)=1-\frac34=\frac14, \] \[ P(B^c)=1-\frac56=\frac16, \] \[ P(C^c)=1-\frac23=\frac13. \]
Step 2: Find the probability that only \(A\) occurs. Since the events are independent, \[ P(A\cap B^c\cap C^c) = P(A)P(B^c)P(C^c). \] \[ = \frac34\cdot\frac16\cdot\frac13. \] \[ = \frac1{24}. \]
Step 3: Find the probability that only \(B\) occurs. \[ P(A^c\cap B\cap C^c) = P(A^c)P(B)P(C^c). \] \[ = \frac14\cdot\frac56\cdot\frac13. \] \[ = \frac5{72}. \]
Step 4: Find the probability that only \(C\) occurs. \[ P(A^c\cap B^c\cap C) = P(A^c)P(B^c)P(C). \] \[ = \frac14\cdot\frac16\cdot\frac23. \] \[ = \frac1{36}. \]
Step 5: Add the three mutually exclusive cases. \[ P(\text{exactly one event occurs}) = \frac1{24} + \frac5{72} + \frac1{36}. \] Taking LCM \(=72\), \[ = \frac3{72} + \frac5{72} + \frac2{72}. \] \[ = \frac{10}{72}. \] \[ = \frac5{36}. \]
Step 6: Write the final answer. \[ \boxed{\frac5{36}} \]
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