From $7$ gentlemen and $4$ ladies, a committee of $5$ is to be formed. The probability that this can be one so as to include at least one lady is

\frac{^{7} C_{5}}{^{11} C_{5}}

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\frac{^{4} C_{1}}{^{11} C_{5}}

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1 - \frac{^{7} C_{5}}{^{11} C_{5}}

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1 - \frac{^{4} C_{1}}{^{11} C_{5}}

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Solution

The correct option is C $1 - \frac{^{7} C_{5}}{^{11} C_{5}}$
Required probability = $1 -$ Probability of no lady being selected = $1 - \frac{^{7} C_{5}}{^{11} C_{5}}$

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