Bunuel wrote:
A playgroup is made up entirely of n pairs of siblings, including the siblings Adam and Josh. 4 members of the playgroup are chosen to represent it in a competition. What is the value of n?
(1) The probability that Adam and Josh are among the 4 members chosen to represent the playgroup is 2/5
(2) The probability that 2 sibling pairs are chosen to represent the playgroup is 1/5
Solution: Pre Analysis:- Total pairs \(= n\)
- Total people \(= 2n\) (including Adam and Josh)
- 4 members are chosen from this 2n people \(^{2n}C_{4}\) ways
- We are asked the value of \(n\)
Statement 1: The probability that Adam and Josh are among the 4 members chosen to represent the playgroup is 2/5
- Since Adam and Josh are already chosen, we just need to choose other 2 members from \(2n-2\) remaining members
- Which can be done in \(^{2n-2}C_{2}\) ways
- Thus according to this statement, \(\frac{^{2n-2}C_{2}}{^{2n}C_{4}}=\frac{2}{5}\)
\(⇒\frac{\frac{(2n-2)!}{(2n-2-2)!\times 2!}}{\frac{(2n)!}{(2n-4)!\times 4!}}=\frac{2}{5}\)
\(⇒\frac{\frac{(2n-2)!}{(2n-4)!\times 2}}{\frac{2n\times (2n-1)\times (2n-2)!}{(2n-4)!\times 24}}=\frac{2}{5}\)
\(⇒\frac{12}{2n(2n-1)}=\frac{2}{5}\)
\(⇒2(4n^2-2n)=60\)
\(⇒8n^2-4n-60=0\) - Since sum of roots is positive and product is negative, we can infer that we will get one negative and one positive value of n from \(8n^2-4n-60=0\)
- Ignoring the negative value of n, we will get the required value of n
- Thus, statement 1 alone is sufficient and we can eliminate options B, C and E
Statement 2: The probability that 2 sibling pairs are chosen to represent the playgroup is 1/5
- Choosing 2 pairs from n pairs \(= ^{n}C_2\)
- Thus according to this statement, \(\frac{^{n}C_{2}}{^{2n}C_{4}}=\frac{1}{5}\)
\(⇒\frac{\frac{(n)!}{(n-2)!\times 2!}}{\frac{(2n)!}{(2n-4)!\times 4!}}=\frac{1}{5}\)
\(⇒\frac{\frac{n\times (n-1)\times (n-2)!}{(n-2)!\times 2}}{\frac{2n\times (2n-1)\times (2n-2)\times (2n-3)\times (2n-4)!}{(2n-4)!\times 24}}=\frac{1}{5}\)
\(⇒\frac{n(n-1)}{2n\times (2n-1)\times (2n-2)\times (2n-3)}=\frac{1}{60}\)
\(⇒\frac{1}{4(2n-1)(2n-3)}=\frac{1}{60}\)
\(⇒4n^2-8n-12=0\) - Since sum of roots is positive and product is negative, we can infer that we will get one negative and one positive value of n from \(4n^2-4n-12=0\)
- Ignoring the negative value of n, we will get the required value of n
- Thus, statement 2 alone is also sufficient
Hence the right answer is
Option D