A committee of two persons is selected from two men and two women. What is the probability that the committee will have two men ?
The total number of persons $=2+2=4 .$ Out of these four person, two can be selected in $^{4} C _{2}$ ways.
Two men can be selected in $^{2} C _{2}$ way.
Hence $P$ (Two men) $=\frac{^{2} C _{2}}{^{4} C _{2}}$ $=\frac{1}{^{4} C _{2}}=\frac{1}{6}$
$3$ numbers are chosen from first $15$ natural numbers, then probability that the numbers are in arithmetic progression
A committee of two persons is selected from two men and two women. What is the probability that the committee will have one man ?
An unbiased die with faces marked $1, 2, 3, 4, 5$ and $6$ is rolled four times. Out of four face values obtained the probability that the minimum face value is not less than $2$ and the maximum face value is not greater than $5$, is
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