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Question

(A) Total number of students $= 50$. The total number of ways to divide $50$ students into two groups of $20$ and $30$ is given by ${}^{50}C_{20} 	im

Vedclass · Accepted Answer

$\frac{25}{49}$

Vedclass · Answer

(A) Total number of students $= 50$. The total number of ways to divide $50$ students into two groups of $20$ and $30$ is given by ${}^{50}C_{20} 	imes {}^{30}C_{30} = {}^{50}C_{20}$.If both $Ram$ and $Rahim$ are in the first group (size $20$),we need to choose $18$ more students from the remaining $48$ students. The number of ways is ${}^{48}C_{18}$.If both $Ram$ and $Rahim$ are in the second group (size $30$),we need to choose $28$ more students from the remaining $48$ students. The number of ways is ${}^{48}C_{28}$.The required probability is $P = \frac{{}^{48}C_{18} + {}^{48}C_{28}}{{}^{50}C_{20}}$.Using the formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:$P = \frac{\frac{48!}{18!30!} + \frac{48!}{28!20!}}{\frac{50!}{20!30!}} = \frac{48!}{18!30!} \cdot \frac{20!30!}{50!} + \frac{48!}{28!20!} \cdot \frac{20!30!}{50!}$$P = \frac{20 	imes 19}{50 	imes 49} + \frac{30 	imes 29}{50 	imes 49} = \frac{380 + 870}{2450} = \frac{1250}{2450} = \frac{25}{49}$.Thus,the correct option is $A$.

From a group of $50$ students,two sections comprising of $20$ and $30$ students are formed. If $Ram$ and $Rahim$ are two particular students among the $50$ students,then the probability that they both belong to the same section is

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