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(D) Let each friend have $x$ daughters.Total number of daughters is $x + x = 2x$.The total number of ways to choose $3$ tickets out of $2x$ daughters

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let each friend have $x$ daughters.Total number of daughters is $x + x = 2x$.The total number of ways to choose $3$ tickets out of $2x$ daughters is $^{2x}C_3$.The number of ways to choose $3$ tickets such that all go to the daughters of $A$ is $^xC_3$.The probability is given by $\frac{^xC_3}{^{2x}C_3} = \frac{1}{20}$.Expanding the combinations: $\frac{x(x-1)(x-2)}{2x(2x-1)(2x-2)} = \frac{1}{20}$.Simplifying: $\frac{x(x-1)(x-2)}{2x(2x-1)2(x-1)} = \frac{1}{20}$.$\frac{x-2}{4(2x-1)} = \frac{1}{20}$.$5(x-2) = 2x-1$.$5x - 10 = 2x - 1$.$3x = 9 \Rightarrow x = 3$.

Two friends $A$ and $B$ have an equal number of daughters. There are three cinema tickets which are to be distributed among the daughters of $A$ and $B$. The probability that all the tickets go to the daughters of $A$ is $1/20$. The number of daughters each of them has is:

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