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 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
$\frac{9}{10}$
$\frac{9}{20}$
$\frac{19}{20}$
$\frac{17}{20}$
Solution
According to the given information, If drawn number from $A$ is 1 , then the favourable drawn number from $B$ are $2,3,4, \ldots \ldots 10$, are total 9 cases.
Similarly, for 2 , there are 8 cases and so on.
$
\begin{aligned}
\therefore \text { Required probability } & =\frac{9+8+7+\ldots+2+1}{10 \times 10} \\
& =\frac{9 \times 5}{10 \times 10}=\frac{9}{20}
\end{aligned}
$
Hence, option (b) is correct