Question

A team of 8 couples (husband &wife;) attend a Lucky draw in which 4 people are picked up for a prize, then the probability that there is at least one couple is. A.$\dfrac{11}{39}$$$$$ B. $\dfrac{12}{39}$
C. \dfrac{14}{39}$$$$$ D. \dfrac{15}{39}$$$$$

Answer 1

We find the total number of outcomes from the number of selections of 4 individuals out of given 16 individuals as $n\left( S \right)$.We find the number of favourable outcomes $n\left( A \right)$ by filling 4 empty spots for selection by 1 couple or 2 couples. If we select 1 couple for the 4 empty spots we have to fill 2 empty spots by either 1 husband 1 wife who is not a couple or 2 husbands or 2 wives. We use the rule of product, rule of sum and the combinatorial formula for selection of $r$ objects from $n$ distinct objects ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We find the probability $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$.$$$$

Complete step-by-step solution
We are given the question that there are 8 couples (husband &wife;). So there are a total of 16 numbers of individuals. We are going to select 4 individuals for the prize and we are asked to find the probability that there are a couple among the selected 4 individuals. We can select 4 individuals out of 16 in ${}^{16}{{C}_{4}}$ ways which the total possible outcomes as
$n\left( S \right)={}^{16}{{C}_{4}}=\dfrac{16!}{4!12!}=\dfrac{16\cdot 15\cdot 13\cdot 12}{4\cdot 3\cdot 2\cdot 1}=1820$
Let us denote the event of getting a couple among the 4 selected individuals as $A$. The event $A$ can occur in 2 ways either there is 1 couple or 2 couples among the selected 4 individuals. $$$$
Case-1: If 1 couple among the 4 selected individuals we can select 1 couple out of 8 couples in ${}^{8}{{C}_{1}}=8$way and fill 2 spots for selection. Now we have the rest $4-2=2$ spots to fill in a way that they are not coupled. We have $8-1=7$ husbands and also $8-1=7$ wives. We can fill the two empty spots with 1 wife and 1 husband or 2 wives or 2 husbands.

H & W & \\_ & \\_ \\\ \end{matrix}$$ If we are going to select one husband and one wife who are not couples we can fill one empty spot by 1 husband (or wife) in ${}^{7}{{C}_{1}}$ ways. Now in order to ensure another couple does not get into our selection, we exclude his wife (or husband) from further selection. Now we can fill the other empty spot with 1 wife from rest $7-1=6$ wives in ${}^{6}{{C}_{1}}$ ways. So we can fill the 2 empty spots with 1 wife and 1 husband in ${}^{7}{{C}_{1}}\times {}^{6}{{C}_{1}}=42$ ways.$$$$ We can select 2 wives to fill 2 empty spots from 7 wives in ${}^{7}{{C}_{2}}=21$ways or we can select 2 husbands to fill 2 empty spots from 7 husbands in ${}^{7}{{C}_{2}}=21$.$$$$ So by the rule of sum, we can fill the two empty spots in $42+21+21=84$ ways, and by the rule of product, we can fill the 4 empty spots with exactly 1 couple in $8\left( 42+21+21 \right)=8\times 84=672$ways.$$$$ Case-2: We fill the 4 empty spots with 2 couples from 8 couples in ${}^{8}{{C}_{2}}=28$ ways. $$\begin{matrix} H & W & H & W \\\ \end{matrix}$$ We can select 4 people for the prize following case-1 or case-2. So by the rule of sum, the number of favorable outcomes is $$n\left( A \right)=672+28=700$$ So the required probability is; $$P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{700}{1820}=\dfrac{5}{13}=\dfrac{15}{39}$$ **So the correct option is D.** **Note:** We note that the selection of individuals is random which means we have not biased towards selection of any particular individuals. We can alternatively solve using negation where we find the $P\left( A \right)=1-P\left( {{A}^{'}} \right)$ where ${{A}^{'}}$ is event of selection no couple. We can find $n\left( {{A}^{'}} \right)$ by selecting 1 husband (or wife ) and rejecting his/her spouse for the next selection 4 times. We can get $n\left( {{A}^{'}} \right)={}^{16}{{C}_{1}}\times {}^{14}{{C}_{1}}\times {}^{12}{{C}_{1}}\times {}^{10}{{C}_{1}}$.