Three groups of children A, B, and C contain | vedclass

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(C) Let $G$ denote a girl and $B$ denote a boy. The probabilities for each group are as follows:Group $A$: $P(G_A) = \frac{3}{4}$, $P(B_A) = \frac{1}{

Vedclass · Accepted Answer

$\frac{13}{32}$

Vedclass · Answer

(C) Let $G$ denote a girl and $B$ denote a boy. The probabilities for each group are as follows:Group $A$: $P(G_A) = \frac{3}{4}$, $P(B_A) = \frac{1}{4}$Group $B$: $P(G_B) = \frac{2}{4}$, $P(B_B) = \frac{2}{4}$Group $C$: $P(G_C) = \frac{1}{4}$, $P(B_C) = \frac{3}{4}$We need to select $1$ girl and $2$ boys. The possible cases are:CaseSelection from $A, B, C$Probability$1$$G, B, B$$\frac{3}{4} 	imes \frac{2}{4} 	imes \frac{3}{4} = \frac{18}{64}$$2$$B, G, B$$\frac{1}{4} 	imes \frac{2}{4} 	imes \frac{3}{4} = \frac{6}{64}$$3$$B, B, G$$\frac{1}{4} 	imes \frac{2}{4} 	imes \frac{1}{4} = \frac{2}{64}$Total probability $= \frac{18}{64} + \frac{6}{64} + \frac{2}{64} = \frac{26}{64} = \frac{13}{32}$.

List-$I$	List-$II$
$(i) \ P(A \cap B)$	$(1) \ 0.4$
$(ii) \ P(A \cap \bar{B})$	$(2) \ 0.2$
$(iii) \ P(\bar{A} \cap B)$	$(3) \ 0.3$
$(iv) \ P(\bar{A} \cap \bar{B})$	$(4) \ 0.1$

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