Question

There is a box containing 33 chits with numbers from 1 to 33 written on it. Yudishter picks up a chit at random from the box and put it back. Duryodhan then picks up a chit from the same box with 33 chits. What is the probability that the number picked by Yudishter is more than Duryodhan?
33/64
16/33
1/2
32/33​

Answer 1

Concept

In this problem, we have to deal with probability. Here, first, we will find the probability of picking up the first chit and then find the conditional probability of occurrence of the second chit when the first chit has been drawn and finally apply the multiplication theorem on probability.

Given

We have given, a box containing $33$ chits with numbers from $1$ to $33$ written on it. Yudishter picks up a chit at random from the box and put it back. Duryodhan then picks up a chit from the same box with $33$ chits.

To Find

We have to find the probability that the number picked by Yudishter is more than Duryodhan.

Solution  

A box containing $33$ chits numbered from   $[{1,2,3,4,...,33}]$

The number picked by Yudishter is more than Duryodhan.

Let the number picked up by Duryodhan be $1$ then the number picked up by Yudishter should be greater than $1$ or Yudishter can pick up a number

in $32$ ways. And Duryodhan can pick a chit in $33$ ways.

Probability = Number of favorable possibilities / Total number of possibilities.

Number of favorable possibilities $=32\times33$

Total number of possibilities $=33\times33$

The probability that the number picked by Yudishter is more than Duryodhan = Number of favorable possibilities / Total number of possibilities

$= \frac{32\times33}{33\times33}$

$=\frac{32}{33}$

As a result, the probability that the number picked by Yudishter is more than Duryodhan is $\frac{32}{33}$