Practicing Success
Practicing Success
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2n boys are randomly divided into two subgroups containing n boys each. The probability that the two tallest boys are in different groups is: |
$\frac{n}{2n-1}$ $\frac{n-1}{2n-1}$ $\frac{2n-1}{4n^2}$ None of these |
$\frac{n}{2n-1}$ |
P(two tallest are in different groups) = $\frac{{^{2n-2}C}_{n-1}.{^2C}_1}{{^{2n}C}_n}=\frac{n}{2n-1}$ |
