Four persons are selected at random out of 3 men, 2 women and 4 childr

1.	Four persons are selected at random out of 3 men, 2 women and 4 children. The probability that there are exactly 2 children in the selection isA. \(\frac{11}{21}\) B. \(\frac{9}{21}\) C. \(\frac{10}{21}\) D. none of these
Answer» As there are 3 + 2 + 4 = 9 persons So, 4 persons out of 9 can be drawn in ⁹C₄ ways = 126 Let E denote the event that there are exactly 2 children in the selection. 2children out of 4 can be selected in ⁴C₂= 6 ways And rest two persons can be male or female. So we will select 2 persons out of remaining 5. ∴ P(E) = \(\frac{4C_2\times5C_2}{126}\) = \(\frac{6\times10}{126}\) = \(\frac{10}{21}\) Hence, P(E) = \(\frac{10}{21}\) As our answer matches only with option (c) ∴ Option (c) is the only correct choice.

Four persons are selected at random out of 3 men, 2 women and 4 children. The probability that there are exactly 2 children in the selection isA. \(\frac{11}{21}\) B. \(\frac{9}{21}\) C. \(\frac{10}{21}\) D. none of these

Answer»

As there are 3 + 2 + 4 = 9 persons

So, 4 persons out of 9 can be drawn in ⁹C₄ ways = 126

Let E denote the event that there are exactly

2 children in the selection. 2children out of 4 can be selected in ⁴C₂= 6 ways

And rest two persons can be male or female. So we will select 2 persons out of remaining 5.

∴ P(E) = \(\frac{4C_2\times5C_2}{126}\) = \(\frac{6\times10}{126}\) = \(\frac{10}{21}\)

Hence,

P(E) = \(\frac{10}{21}\)

As our answer matches only with option (c)

∴ Option (c) is the only correct choice.

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