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Mathematics Question on Conditional Probability

Assume that in a family, each child is equally likely to be a boy or a girl. AA family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is

A

12\frac{1}{2}

B

13\frac{1}{3}

C

23\frac{2}{3}

D

47\frac{4}{7}

Answer

47\frac{4}{7}

Explanation

Solution

Sample space, S=BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGGS = \\{BBB, \,BBG,\, BGB, \,GBB,\, GGB, \,GBG,\, BGG,\, GGG\\} Let E1E_1 be the event that eldest child is a girl and E2E_2 be the event that atleast one child is a girl. E1=GBB, GGB, GBG, GGGE_1 = \\{GBB,\, GGB,\, GBG,\, GGG\\} P(E1)=48P\left(E_{1}\right) = \frac{4}{8} E_{ 2} = \left\\{BBG,\, BGB,\, GBB,\, GGB, \,GBG,\, BGG,\, GGG\right\\} P(E2)=78P\left(E_{2}\right) = \frac{7}{8} E_{1} \cap E_{ 2} = \left\\{GBB,\, GGB, \,GBG, \,GGG\right\\} P(E1∩E2)=48P\left(E_{1} \cap E_{ 2}\right) = \frac{4}{8} ∴\therefore Required probability =P(E1∣E2)=P(E1∩E2)P(E2)= P \left(E_{1} | E_{2} \right) =\frac{P\left(E_{1} \cap E_{ 2}\right)}{P\left(E_{2}\right)} =4/87/8=47= \frac{4/8 }{7/8 } = \frac{4}{7}