Question:
A husband and wife appeared for an interview for two vacancies in an office. The probability of the husband's selection is \(\frac{1}{5}\) and that of the wife's selection is \(\frac{1}{7}\), then the probability that exactly one of them is selected is...
A husband and wife appeared for an interview for two vacancies in an office. The probability of the husband's selection is \(\frac{1}{5}\) and that of the wife's selection is \(\frac{1}{7}\), then the probability that exactly one of them is selected is...
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Add the probability that only the husband is selected to the probability that only the wife is selected, treating the events as independent.
Updated On: Nov 12, 2026
- \(\frac{1}{35}\)
- \(\frac{11}{35}\)
- \(\frac{2}{7}\)
- \(\frac{1}{7}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
The selections of the husband and the wife are independent events.
Let \(P(H)=\dfrac15\) and \(P(W)=\dfrac17\).
Step 2: Exactly One
\(P=P(H)P(\bar W)+P(\bar H)P(W)\).
Step 3: Substitute
\(P=\dfrac15\cdot\dfrac67+\dfrac45\cdot\dfrac17=\dfrac{6}{35}+\dfrac{4}{35}\).
Step 4: Simplify
\(P=\dfrac{10}{35}=\dfrac27\).
Final Answer:
The probability is 2 over 7. \[ \boxed{\text{(C) }\dfrac{2}{7}} \]
The selections of the husband and the wife are independent events.
Let \(P(H)=\dfrac15\) and \(P(W)=\dfrac17\).
Step 2: Exactly One
\(P=P(H)P(\bar W)+P(\bar H)P(W)\).
Step 3: Substitute
\(P=\dfrac15\cdot\dfrac67+\dfrac45\cdot\dfrac17=\dfrac{6}{35}+\dfrac{4}{35}\).
Step 4: Simplify
\(P=\dfrac{10}{35}=\dfrac27\).
Final Answer:
The probability is 2 over 7. \[ \boxed{\text{(C) }\dfrac{2}{7}} \]
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