Question

A has 3 shares in a lottery in which there are 3 prizes and 6 blanks; B has 1 share in a lottery in which there is 1 prize and 2 blanks; show that A’s chance of success is to B’s as 16 to 7.

Answer 1

Hint:Success means getting a prize on any of the shares in the lottery. Calculate sample space for each case by selecting total number of shares from total tickets. Calculate total number of ways of not getting any prize in shares given from the tickets. Now get success probability by subtracting it by 1. Use $C_{r}^{n}$ for selecting 'r' items out of ‘n’, using the formula, $C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$ .

Complete step-by-step answer:
Here, it is given that A has 3 shares in a lottery and B has 1 share in which there are 3 prizes and 6 blanks for A and 1 prize and 2 blanks for B. And we need to determine the ratio of chance of success for A and chances of success for B.
Now, there is a chance of success of A and B if they select a share of any prize in the lottery.
So, first let us calculate the chance of success with respect to 3 shares in a lottery and in which he has 3 prizes and 6 blanks.
As we know that number of ways of selecting ‘r’ items from n items is given as:
$C_{r}^{n}$
Now, total number ways of selection 3 shares from total (3+6) or 9 tickets;
$=C_{3}^{9}$
Number of ways of selecting 3 shares from 6 blanks, we get
$=C_{3}^{6}$
Now, probability of any event is given as
$P(E)=\dfrac{\text{Favourable cases}}{\text{Total cases}}$
So, chance of losing a prize from the given lottery can be given as
$P=\dfrac{\text{Number of ways of blanks lottery}}{\text{Total cases of selection of 3 shares}}$
$P=\dfrac{C_{3}^{6}}{C_{3}^{9}}$
As we know that summation of occurring of an event and non-occurring is 1.
Hence, chances of success for A is given as
$P(A)=1-P=1-\dfrac{C_{3}^{6}}{C_{3}^{9}}.............\left( i \right)$
Similarly, number of selecting 1 share from total 1+2 = 3 tickets of lottery is given as $C_{1}^{3}$
And the number of selecting 1 share from 2 blanks be $C_{1}^{2}$ .
Hence, probability of failure of B in lottery can be given as,
$P=\dfrac{C_{1}^{2}}{C_{1}^{3}}$
Hence, chances of success for B can be given as,
$1-P=1-\dfrac{C_{1}^{2}}{C_{1}^{3}}$ or
$P\left( B \right)=1-\dfrac{C_{1}^{2}}{C_{1}^{3}}...............\left( ii \right)$
Hence, equation (i) can be simplified as
$P(A)==1-\dfrac{\dfrac{6!}{3!3!}}{\dfrac{9!}{6!3!}}$
Hence,
$P(A)=1-\dfrac{\dfrac{6\times 5\times 4}{6}}{\dfrac{9\times 8\times 7}{6}}$
$P(A)=1-\dfrac{20}{84}=\dfrac{64}{84}=\dfrac{16}{21}$
Hence, we get
$P(A)=\dfrac{16}{21}...........(iii)$
Now, similarly we can simplify equation (ii) we get,
$P(B)=1-\dfrac{2}{3}=\dfrac{1}{3}$
$P(B)=\dfrac{1}{3}...........(iv)$
Therefore, the chance of success of A is $\dfrac{16}{21}$ and chances of success of B is $\dfrac{1}{3}$. So, the ratio of both can be given as,
 $\begin{align}
  & R=\dfrac{\left( \dfrac{16}{21} \right)}{\left( \dfrac{1}{3} \right)}=\dfrac{16}{21}\times \dfrac{3}{1} \\
 & \text{R=}\dfrac{16}{7} \\
\end{align}$
Hence, A’s chance of success is to B’s as 16 to 7.

Note: Another approach for calculation of success would be that,
Chance of success of $A=\dfrac{C_{1}^{3}C_{2}^{6}+C_{2}^{3}C_{1}^{6}+C_{3}^{3}C_{0}^{6}}{C_{3}^{9}}$ and
Chance of success of \[B=\dfrac{C_{1}^{1}C_{0}^{2}}{C_{1}^{3}}\]
One can go wrong while selecting ‘r’ items from ‘n’ items. He or she may apply $P_{r}^{n}$ identity in place of $C_{r}^{n}$ . Be careful in selection and permutation or arrangements.