Question

A has 3 shares in the lottery containing 3 prizes and 9 blanks. B has two shares in a lottery containing 2 prizes and 6 blanks; compare their chance of success.
A) 715: 952
B) 952: 715
C) 237: 715
D) 952: 237

Answer 1

Hint:
In this question we have to compare two draws and their chances of success.
Total number of tickets with A is 9+3 = 12 means P(n) = 12
Out of the total tickets, A took 3 tickets
Then, Total P(n) = $^{12}{C_3}$
We will simply solve using factorial methods.
Now, we will see favorable case:
(W, L, L) + (W, W, L) + (W, W, W)
Where, W = win, L = lose
Then, $3{C_1} \times 9{C_2} + 3{C_2} \times 9{C_1} + 3{C_3}$
We will easily solve this process.
As we know that.
Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes.
P(A) = $\dfrac{{The\,number\,of\,wanted\,outcome}}{{The\,number\,of\,possible\,outcomes}}$

Complete step by step solution:
Since A has 3 shares in a lottery, his chance of success means that he gets at least one prize, that is, he gets either one prize or 2 prizes or 3 prizes and chance of failure means that he gets no prize.
It is certain that either he succeeds or fails. If ‘P’ denotes his chance of success and ‘q’ the chance of his failure, then;
$P + q = 1$ or $P = 1 – q$
Now, q x n = Total number ways
${ \Rightarrow ^{12}}{C_3} = \dfrac{{12 \times 11 \times 10}}{{1 \times 2 \times 3}}$
$ \Rightarrow 220$
Since out of 12 tickets in the lottery, he can draw any 3 tickets by virtue of his having 3 shares in the lottery and M = Favorable number of ways,
${ \Rightarrow ^9}{C_3} = \dfrac{{9 \times 8 \times 7}}{{1 \times 2 \times 3}}$
= 84
Since, he will fail to draw a prize if all the tickets drawn by him are blanks.
Therefore, q = $\dfrac{m}{n} = \dfrac{{Total\,number\,of\,ways}}{{Favorable\,number\,of\,ways}}$
$ \Rightarrow \dfrac{{84}}{{220}} = \dfrac{{21}}{{55}}$
Therefore, P = A’s chance of success = $1 - \dfrac{{21}}{{55}}$
$ \Rightarrow \dfrac{{55 - 21}}{{55}} = \dfrac{{34}}{{55}}$
Similarly, B’s chance of success:
$ \Rightarrow {P^1} = 1 - {q^1}$
$ = \dfrac{{1{ - ^6}{C_2}}}{{^8{C_2}}}$
$ = 1 - \dfrac{{6 \times 5}}{{8 \times 7}} = 1 - \dfrac{{15}}{{28}}$
$ = 1 - \dfrac{{15}}{{28}} = \dfrac{{28 - 15}}{{28}} = \dfrac{{13}}{{28}}$
Thus, ${P^1} = \dfrac{{13}}{{28}}$
Therefore, A’s chance of success = B’s chance of success
$ \Rightarrow P:{P^1} = \dfrac{{34}}{{55}}:\dfrac{{13}}{{28}}$(using cross multiplication)

$ \Rightarrow 952: 715$

Note:
In this question we have used the concept of probability as well as combination. You can see that here we have assumed P is the chance of success and q is the chance of failure. We have written $P + q = 1$ because the sum of probability is 1.