Question

In a lottery $50$ tickets are sold in which \[14\] are prizes. A man bought $2$ tickets, then the probability that the man win the prize is 

A. $\dfrac{{17}}{{35}}$ 

B. $\dfrac{{18}}{{35}}$ 

C. $\dfrac{{72}}{{175}}$ 

D. $\dfrac{{13}}{{175}}$ 

Answer 1

Hint: First, we shall analyze the given information so that we are able to solve the problem. Here, we are given some data. We are asked to calculate the probability of a man getting two tickets that will receive prizes.

 We need to apply the formula listed below to obtain the desired probability.

Formula used:

a) The formula to calculate the probability of an event is as follows.

The probability of an event (say A),$P\left( A \right) = \dfrac{\text{Number of favourable outcomes}}{\text{Number of total outcomes}}$

b) The formula when the events are opposite to each other is as follows. 

$P\left( A \right) + P\left( {{A^{ - 1}}} \right) = 1$ 

Complete step by step answer:

It is given that there are fifty lottery tickets.

Hence, the total number of tickets $ = 50$ 

Also, it is given that \[14\] tickets will receive prizes out of fifty tickets.

Hence, there will be $50 - 14 = 36$ tickets that will not receive prizes.

Here, a man buys two tickets.

He bought two tickets from fifty tickets.

Hence, the number of ways he can buy two tickets from fifty tickets${ = ^{50}}{C_2}$ 

$^{50}{C_2} = \dfrac{{50!}}{{2!\left( {50 - 2} \right)!}}$    (Here we applied the combination formula)

$ = \dfrac{{50 \times 49 \times 48!}}{{2!48!}}$    

$ = 25 \times 49$ 

$ = 1225$ 

Also, the number of ways he can buy two tickets that will not receive a prize (from$36$tickets)${ = ^{36}}{C_2}$

\[^{36}{C_2} = \dfrac{{36!}}{{2!\left( {36 - 2} \right)!}}\]    (Here we applied the combination formula)

$ = \dfrac{{36 \times 35 \times 34!}}{{2!34!}}$    

$ = 18 \times 35$ 

$ = 630$ 

Now, we shall find the probability of a man getting two tickets that will not receive prizes.

We shall apply the probability of an event.

The formula to calculate the probability of an event is as follows.

The probability of an event (say A),$P\left( A \right) =\dfrac{\text{Number of favourable outcomes}}{\text{Number of total outcomes}}$

Let \[P\left( A \right)\] be the probability of a man getting two tickets that will not receive prizes.

Then the probability of a man getting two tickets that will not receive prizes$P\left( A \right) = \dfrac{{^{36}{C_2}}}{{^{50}{C_2}}}$ 

\[P\left( A \right) = \dfrac{{630}}{{1225}}\] 

\[ \Rightarrow P\left( A \right) = \dfrac{{18}}{{35}}\] 

We are asked to calculate the probability of a man getting two tickets that will receive prizes.

Let \[P\left( {A'} \right)\] be the probability of a man getting two tickets that will receive prizes.

Now, we shall apply the complimentary formula.

\[P\left( {A'} \right) = 1 - P\left( A \right)\] 

\[ = 1 - \dfrac{{18}}{{35}}\] 

\[ = \dfrac{{35 - 18}}{{35}}\] 

\[ = \dfrac{{17}}{{35}}\] 

Hence the required probability is\[\dfrac{{17}}{{35}}\]

So, the correct answer is “Option A”.

Note: Alternative way:- We can take the event as getting a prize then we will have two cases.

Case-1 Man gets one prize ticket and one without prize. So for this case we will get favourable outcomes as-

$=^{14}C_1 \times ^{36}C_1$ = $14\times 36$ 

$= 504$

Case-2 Man gets prizes in both tickets. So for this case we will get favourable outcomes as-

$=^{14}C_2 $=$\dfrac{14 \times 13 \times 12!}{2! \times 12!}$

$=91$

So total favourable outcomes = $504+91$

$=595$

So using probability formula-

Probability that the man win the prize = $\dfrac{\text{Number of favourable outcomes}}{\text{Number of total outcomes}}$

$=\dfrac{595}{1225}$

$=\dfrac{17}{35}$

We apply the complementary probability formula when the two events are opposite to each other.