There are four men and six women on the city councils. If one council member is selected for the committee at random, how likely that it is a woman?
Answer
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Hint:
There are 4 men and 6 women in the city councils. We first find the number of ways of choosing 1 person from 10 people and then we find the number of ways of choosing 1 woman out of 6 women, then we find the probability of selecting 1 woman from 10 people.
Complete step by step solution:
We have 4 men and 6 women in the city councils.
Out of 4 men and 6 women one person can be chosen in ,\[{}^{10}{C_1} = 10\;\]ways.
As x objects can be chosen from y objects in \[{}^y{C_x}\;\]ways,
And again,
The number of ways of selecting 1 women out of 6 women \[ = \;{}^6{C_1} = 6\].
Therefore, Required probability \[ = \dfrac{6}{{10}}\,\, = \dfrac{3}{5}\]
Note:
There are many problems in which we are interested in determining the number of ways in which k objects can be selected from n distinct objects without regard to the order in which they are selected. Such selections are called combinations of k-sets. It may help to think of combinations as a committee.
There are 4 men and 6 women in the city councils. We first find the number of ways of choosing 1 person from 10 people and then we find the number of ways of choosing 1 woman out of 6 women, then we find the probability of selecting 1 woman from 10 people.
Complete step by step solution:
We have 4 men and 6 women in the city councils.
Out of 4 men and 6 women one person can be chosen in ,\[{}^{10}{C_1} = 10\;\]ways.
As x objects can be chosen from y objects in \[{}^y{C_x}\;\]ways,
And again,
The number of ways of selecting 1 women out of 6 women \[ = \;{}^6{C_1} = 6\].
Therefore, Required probability \[ = \dfrac{6}{{10}}\,\, = \dfrac{3}{5}\]
Note:
There are many problems in which we are interested in determining the number of ways in which k objects can be selected from n distinct objects without regard to the order in which they are selected. Such selections are called combinations of k-sets. It may help to think of combinations as a committee.
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