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Question: A person has \[3\] shares in a lottery containing \[2\] prizes and \[8\] blanks. The chance of getti...

A person has 33 shares in a lottery containing 22 prizes and 88 blanks. The chance of getting prices is:
A. 1228\dfrac{{12}}{{28}}
B. 38\dfrac{3}{8}
C. 715\dfrac{7}{{15}}
D. 815\dfrac{8}{{15}}

Explanation

Solution

In order to solve the question given above, we will use the method of combinations. It refers to the grouping of items and in combinations the order does not matter. We will be using a very simple formula to calculate the number of ways to choose the ticket, which is, C(n,r)C\left( {n,r} \right) or nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}} .

Formula used:
While solving this question we will be using the formula of calculating combinations:
nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}} .

Complete step by step solution:
We are given that the man has 33 shares in a lottery and the lottery has 22 prizes and 88 blanks.
The probabilities of the man winning are: WLL, WWL, WWW.
The first probability: WLL, where the man has one chance of winning and two chances of losing =3C1×9C2{ = ^3}{C_1}{ \times ^9}{C_2} .
The second probability: WWL, where the man has two chances of getting the price and 1 chance of losing =3C2×9C1{ = ^3}{C_2}{ \times ^9}{C_1} .
The third probability: WWW, where the man has three chances of getting the price and no chance of losing =3C3×9C0{ = ^3}{C_3}{ \times ^9}{C_0} .
Now, we have to calculate the total number of ways in which the man can choose 33 out of the total of 1212 tickets =12C3{ = ^{12}}{C_3} .
Now, we find the probability of him getting the prize.
P(A)=(3C1×9C2+3C2×9C1+3C3×9C0)12C0P\left( A \right) = \dfrac{{\left( {^3{C_1}{ \times ^9}{C_2}{ + ^3}{C_2}{ \times ^9}{C_1}{ + ^3}{C_3}{ \times ^9}{C_0}} \right)}}{{^{12}{C_0}}}
=3455= \dfrac{{34}}{{55}} . (solved using the formula given above)
And, P(B)=(2C2×6C0+2C1×6C1)8C2P\left( B \right) = \dfrac{{\left( {^2{C_2}{ \times ^6}{C_0}{ + ^2}{C_1}{ \times ^6}{C_1}} \right)}}{{^8{C_2}}}
=1228= \dfrac{{12}}{{28}} .

So, the correct answer is Option A.

Note: Always remember that while solving the sums similar to one given above use the method of probability and combinations. To calculate the values of the combinations, we use the formula that has been already discussed above in the answer. nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}} , remember it while solving such sums.