A committee of five is to be chosen from a group of $9$ people. The probability that a certain married couple will either served together or not at all is:

\frac{1}{2}

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\frac{5}{9}

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\frac{4}{9}

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\frac{2}{3}

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Solution

The correct option is C $\frac{4}{9}$
Total no of ways to choose $5$ people out of $9 =^{9} C_{5} = 126$
Couple serves committee $=^{7} C_{3} = 35$
Couple do not serves committee $^{7} C_{5} = 21$
Couple together serve or do not serve
So favourable ways $= 21 + 35 = 56$
Probability $= \frac{56}{126} = \frac{4}{9}$

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[CEE 1993]

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A committee of five is to be chosen from a group of 9 people. The probability that a certain married couple will either served together or not at all is:

The correct option is C 49Total no of ways to choose 5 people out of 9=9C5=126Couple serves committee =7C3=35 Couple do not serves committee 7C5=21Couple together serve or do not serve So favourable ways =21+35=56Probability =56126=49

A committee of five is to be chosen from a group of $9$ people. The probability that a certain married couple will either served together or not at all is: