Question

A box contains $25$ tickets number $1,2,....25$. If two tickets are drawn at random then the probability that the product of their numbers is even, is

• A. $\frac{11}{50}$
• B. $\frac{13}{50}$
• C. $\frac{37}{50}$
• D. $Noneofthese$

Answer 1

The correct option is C $\frac{37}{50}$

We have,

Total number of tickets with numbers$=25$

Number of tickets with even numbers on them$=12$

Number of tickets with odd numbers on them$=13$

Number of tickets drawn$=2$

Now,

Selecting 2 tickets from the 25

Total number of possible choices$=$ number of ways in which 2 tickets can be drawn from the total $25$

So,

$n{=}^{25}{C}_2$

$=\frac{25\times 24}{2\times 1}$

$=300$

Product of two numbers would be even

If at least one of the numbers is even

So,

Number of ways in which 2 tickets with even numbers on both of them can be drawn + number of ways in which 2 tickets with an even number on one of them and an odd number an another can be drawn

$m_A=m_{E{A}_1}+m_{E{A}_2}$

$=\left({}^{12}{C}_2\right)+\left({}^{13}{C}_1{\times }^{12}{C}_1\right)$

$=\frac{12\times 11}{2\times 1}+\frac{12}{1}\times \frac{13}{1}$

$=(6\times 11)+(12\times 13)$

$=66+156$

$=222$

Then,

Probability

$P\left(A\right)=\frac{m_A}{n}$

$=\frac{222}{300}$

$=\frac{37}{50}$

Hence, this is the answer.