Question

Three small sized Herds A, B and C consist of $3$ cows and $1$ buffalo, $2$ cows and $2$ buffaloes, $1$ cow and $3$ buffaloes, respectively. Find the probability of selecting one cow and two buffaloes from three Herds.

• A. $\frac{3}{32}$
• B. $\frac{13}{32}$
• C. $\frac{17}{32}$
• D. None of these

Answer 1

The correct option is B $\frac{13}{32}$

From the question,

Herd A : $3$ cows, $1$ buffalo

Herd B : $2$ cows, $2$ buffalo

Herd C : $1$ cows, $3$ buffalo

There are may possible way for selecting one cow and $2$ buffalo from the given three herds.

From herd A one cow can be selected out of three $={}^3{C}_1$

and one buffalo $={}^1{C}_1$

$\Rightarrow$ From A $=\frac{{}^3{C}_1{}^1{C}_1}{{}^4{C}_2}=\frac{3}{6}$

Similarly from herd B $=\frac{{}^2{C}_1{}^2{C}_2}{{}^4{C}_3}=\frac{2}{4}$

From herd C $=\frac{{}^1{C}_1{}^3{C}_2}{{}^4{C}_3}=\frac{3}{4}$

This is care $1$

i.e, $\frac{3}{6}\times \frac{2}{4}\times \frac{3}{4}=\frac{6}{32}$

As herd A has one buffalo. One can be selected from A and another from herd B or C.

$\Rightarrow \frac{{}^3{C}_1{}^1{C}_1}{{}^4{C}_2}+\frac{{}^2{C}_1}{{}^4{C}_1}+\frac{{}^3{C}_1}{{}^4{C}_1}=\frac{3}{6}\times \frac{2}{4}\times \frac{3}{4}=\frac{6}{32}$

$\therefore$ Probability $=\frac{6}{32}+\frac{6}{32}+\frac{1}{32}=\frac{13}{32}.$

Hence, the answer is $\frac{13}{32}.$