Question

A committee of two persons is selected from three men and two women. What is the probability that the committee will have a) no man? b) one woman? C) two men?

Answer 1

Hint : Here we are going to use the concept of probability and the combinations. We will make the combinations for the statement which is given in the question statement and then by selecting the right combination.

Complete step by step solution:
The committee consists of three women and two men.
Total number of persons in committee $= 5$
Probability is the state of being probable and the extent to which something is likely to happen in the particular situations or the favourable outcomes. Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the outcomes.
$P(A) =$ Total number of the favourable outcomes / Total number of the outcomes
A.Probability that the committee have no man –
$P(no{\text{ man) = }}\dfrac{{{}^2{C_1}}}{{{}^5{C_2}}} = \dfrac{2}{{10}} = \dfrac{1}{5}$
B.Probability that the committee have one woman –
$P(one{\text{ woman) = }}\dfrac{{{}^3{C_1} \times {}^2{C_1}}}{{{}^5{C_2}}} = \dfrac{{3 \times 2}}{{10}} = \dfrac{3}{5}$
C. The probability that the committee have two men –
$P(two{\text{ men) = }}\dfrac{{{}^2{C_2}}}{{{}^5{C_2}}} = \dfrac{1}{{5}}$

Note: While simplification identify the sequence followed and apply formula accordingly, whether it is arithmetic progression or the geometric progression. The probability of any event always ranges between zero and one. It can never be the negative number or the number greater than one. The probability of impossible events is always equal to zero whereas, the probability of the sure event is always equal to one. Combinations are used if the certain objects are to be arranged in such a way that the order of objects is not important. $^nc{}_r = \dfrac{{n!}}{{r!(n - r)!}}$