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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $= \frac{18}{64} + \frac{6}{64} + \frac{2}{64} = \frac{26}{64} = \frac{13}{32}$.

Similar Questions

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three persons apply for the same house is

If $A, B, C$ are three events of a sample space such that $P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$, then which of the following is correct?

The probability of happening an event $A$ is $0.5$ and that of $B$ is $0.3$. If $A$ and $B$ are mutually exclusive events,then the probability of happening neither $A$ nor $B$ is

What is the probability of having $53$ Sundays in a non-leap year?

Vedclass Test Series

Exam Paper Generator

Online Exam Module