Question

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) ²ⁿC_n

Answer 1

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

Answer 2

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

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