Question

If 7 men 5 women have applied for job and 3 applicants are randomly selected from the probability that 2 are men how many WOmen?
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Answer 1

The probability of selecting 2 men and a woman is  $\frac{21}{44}$

The sample space is made up of 7 men and 5 women. So we have a total of 12 people.

If 3 applicants are randomly selected without replacement, there will be three mutually exclusive possibilities;

$MMW=\text{Event that a Man, then Man, then Woman is selected}\\MWM=\text{Event that a Man, then Woman, then Man is selected}\\WMM=\text{Event that a Woman, then Man, then Man is selected}$

The final probability will have the form

$P(\text{2 men and 1 woman})=P(MMW)+P(MWM)+P(WMM)$

because the possibilities are mutually exclusive.

Computing the Probability of each possibility

Each mutually exclusive possibility is made up of dependent events. This is because when selection is done without replacement, it affects the size of the sample space.

• The Probability of selecting a Man, then a Man, then a Woman is

        $P(MMW)=\dfrac{7}{12}\times\dfrac{6}{11}\times\dfrac{5}{10}\\\\=\dfrac{7}{44}$

• The Probability of selecting a Man, then a Woman, then a Man is

        $P(MWM)=\dfrac{7}{12}\times\dfrac{5}{11}\times\dfrac{6}{10}\\\\=\dfrac{7}{44}$

• The Probability of selecting a Woman, then a Man, then a Man is

        $P(MWM)=\dfrac{5}{12}\times\dfrac{7}{11}\times\dfrac{6}{10}\\\\=\dfrac{7}{44}$

Calculate the Probability of getting two men and a woman

The final probability is

$P(\text{2 men and 1 woman})=P(MMW)+P(MWM)+P(WMM)\\\\=\dfrac{7}{44}+\dfrac{7}{44}+\dfrac{7}{44}\\\\=\dfrac{21}{44}$

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