Question

A box contains 25 tickets number 1,2,………….25. if two tickets are drawn at random then the probability that the product of the number is even is
A.$\dfrac{{11}}{{50}}$
B.$\dfrac{{13}}{{50}}$
C.$\dfrac{{37}}{{50}}$
D.None of these

Answer 1

Hint: We have to calculate the probability of the two tickets, a product of whose number is an even number. Since even numbers are numbers which are divisible by 2.
So, the total ticket with an even number is 12. Firstly we calculate the number of ways in which two tickets can be drawn from 25 tickets. Then we calculate the required probability.

Complete step-by-step answer:
We have the total number of tickets with the number from 1 to 25 = 25.
The number of tickets with even numbers on them that is the number s which are visible by 2 are 12.
The total number of possible choices = the number of ways in which two tickets can be drawn from the total 25 tickets.
So,
$n{ = ^{25}}{C_2} = \dfrac{{25!}}{{2! \times (25 - 2)!}} = \dfrac{{25!}}{{2! \times 23!}}$
$ = \dfrac{{25 \times 24 \times 23!}}{{2 \times 23!}} = \dfrac{{25 \times 24}}{2} = 300$
∴ n = 300
Now, a product of two numbers would be even if at least one of the number from these two number is even. So, the number of ways in which 2 tickets with even numbers on both of them can be draw+ number of ways in which 2 tickets with an even number on one of them and an odd number on another can be drawn.
∴ ${m_A} = {m_{E{A_1}}} + {m_{E{A_2}}}$
Here ${m_A}$ represents the total number even outcomes with the product of both numbers. ${m_{E{A_1}}}$ represents both tickets drawn. ${m_{E{A_2}}}$represents that the ticket is even and the other is odd.
${m_A} = {(^{12}}{C_2}) + {(^{13}}{C_1}{ \times ^{12}}{C_1})$
$ = \dfrac{{12!}}{{2!(12 - 2)!}} + \dfrac{{13!}}{{1!(13 - 1)!}} \times \dfrac{{12!}}{{1!(12 - 1)!}}$
$ = \dfrac{{12 \times 11 \times 10!}}{{2 \times 1 \times 10!}} + \dfrac{{13 \times 12!}}{{1 \times 12!}} \times \dfrac{{12 \times 11!}}{{1 \times 11!}}$
$ = \dfrac{{12 \times 11}}{2} + 13 \times 12$
$ = 66 + 156 = 222$
Therefore the probability is ;
$p(A) = \dfrac{{{m_A}}}{n} = \dfrac{{222}}{{300}} = \dfrac{{37}}{{50}}$. This is the required answer.
That is the probability that the product of the two number on the ticket is even =37/50
So, option (C) is correct.


Note: Here $^n{C_r}$ represent the combination of n things taken r at a time.
Combination:- Each of the different groups which can be formed by taking some or all of the number of objects irrespective of their arrangement is called combination.