There are four men and six women on the city councils. If one council member is selected for a committee at random, how likely is that it is a women?

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Solution

Let E be the event that woman is selected.

$∴ n (E) = 6$

Number of sample space n(S)= Total number of persons in the council = 4 + 6 = 10

$∴ P = \frac{n (E)}{n (S)} = \frac{6}{10} = \frac{3}{5}$

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There are four men and six women on the city councils. If one council member is selected for a committee at random, how likely is that it is a women?

Let E be the event that woman is selected. ∴n(E)=6 Number of sample space n(S)= Total number of persons in the council = 4 + 6 = 10 ∴P=n(E)n(S)=610=35

Let E be the event that woman is selected.

$∴ n (E) = 6$

Number of sample space n(S)= Total number of persons in the council = 4 + 6 = 10

$∴ P = \frac{n (E)}{n (S)} = \frac{6}{10} = \frac{3}{5}$