Each of the two boxes A and B contain 10 chit | vedclass

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(B) Let $X$ be the number drawn from box $A$ and $Y$ be the number drawn from box $B$. The total number of possible outcomes is $10 	imes 10 = 100$.W

Vedclass · Accepted Answer

$\frac{9}{20}$

Vedclass · Answer

(B) Let $X$ be the number drawn from box $A$ and $Y$ be the number drawn from box $B$. The total number of possible outcomes is $10 	imes 10 = 100$.We want to find the probability $P(X The total outcomes can be divided into three cases: $X  Y$,and $X = Y$.The number of cases where $X = Y$ is $10$ (i.e.,$(1,1), (2,2), \ldots, (10,10)$).Since the situation is symmetric,the number of cases where $X  Y$.Let $N$ be the number of cases where $X $2N = 90 \implies N = 45$.The required probability is $\frac{N}{100} = \frac{45}{100} = \frac{9}{20}$.Hence,option $B$ is correct.

Each of the two boxes $A$ and $B$ contain $10$ chits numbered $1$ to $10$. If one chit is drawn at random from each of $A$ and $B$,then the probability that the number on the chit drawn from $A$ is smaller than the number on the chit drawn from $B$,is

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