A group of 2n boys and 2n girls is divided at random into two equal ba

1.	A group of 2n boys and 2n girls is divided at random into two equal batches. The probability that each batch will have equal number of boys and girls is(a) \(\frac{1}{2}n\)(b) \(\frac{1}{4}n\)(c) \(\frac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)d) 2nCn
Answer» (c) \(\frac{(^{2n}C_n)^2}{^{4n}C_{2n}}\) Total number of boys and girls = 2n + 2n = 4n Since, there are two equal batches, each batch has 2n members ∴ Let S (Sample space) : Selecting one batch out of 2 ⇒ S : Selecting 2n members out of 4n members. ⇒ n(S) = ⁴ⁿC_2n If each batch has to have equal number of boys and girls, each batch should have n boys and n girls. Let E : Event that each batch has ‘n’ boys and ‘n’ girls ⇒ n(E) = ²ⁿC_n × ²ⁿC_n = (²ⁿC_n)² ∴ Required probability = \(\frac{n(E)}{n(S)}\) = \(\frac{(^{2n}C_n)^2}{^{4n}C_{2n}}\) Total number of ways of choosing a group is: ⁴ⁿC_n The number of ways in which each group contains equal number of boys and girls is: (²ⁿC_n)(²ⁿC_n) ∴ Required probability is (²ⁿC_n)² / ⁴ⁿC_n optoin c is correct HOPE IT HELPS #IRKI♥︎

A group of 2n boys and 2n girls is divided at random into two equal batches. The probability that each batch will have equal number of boys and girls is(a) \(\frac{1}{2}n\)(b) \(\frac{1}{4}n\)(c) \(\frac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)d) 2nCn

Answer»

(c) \(\frac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)

Total number of boys and girls = 2n + 2n = 4n

Since, there are two equal batches, each batch has 2n members

∴ Let S (Sample space) : Selecting one batch out of 2

⇒ S : Selecting 2n members out of 4n members.

⇒ n(S) = ⁴ⁿC_2n

If each batch has to have equal number of boys and girls, each batch should have n boys and n girls.

Let E : Event that each batch has ‘n’ boys and ‘n’ girls

⇒ n(E) = ²ⁿC_n × ²ⁿC_n = (²ⁿC_n)²

∴ Required probability = \(\frac{n(E)}{n(S)}\) = \(\frac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)

Total number of ways of choosing a group is: ⁴ⁿC_n

The number of ways in which each group contains equal number of boys and girls is:

(²ⁿC_n)(²ⁿC_n)

∴ Required probability is (²ⁿC_n)² / ⁴ⁿC_n

optoin c is correct

HOPE IT HELPS

#IRKI♥︎