From a group of 3 mathematicians, 4 statisticians and 5 economists, a committee of 4 is selected by lottery. The probability that the committee consists of 2 mathematicians and 2 statisticians is

\frac{^{13} C_{2}}{^{12} C_{4}}

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\frac{^{4} C_{2}}{^{12} C_{4}}

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\frac{^{3} C_{2} \times^{4} C_{2}}{^{12} C_{4}}

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\frac{^{7} C_{4}}{^{12} C_{4}}

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Solution

The correct option is C $\frac{^{3} C_{2} \times^{4} C_{2}}{^{12} C_{4}}$
Total number of ways of selecting the committee is $^{12} C_{4}$
The total ways of having 2 mathematicians is to choose the 2 among them. Hence $^{3} C_{2}$
The total ways of having 2 statisticians is to choose the 2 among them. Hence $^{4} C_{2}$
Since, we want both these cases (2 mathematicians AND 2 statisticians) , we use the multiplication principle
The probability is given by $\frac{^{3} C_{2} .^{4} C_{2}}{^{12} C_{4}}$

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