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Mars2501 [29]
2 years ago
6

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?

SAT
1 answer:
ElenaW [278]2 years ago
3 0

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.

<h3>Computing the Probability of each possibility</h3>

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}

<h3>Calculate the Probability of getting two men and a woman</h3>

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}

Learn more about probability here brainly.com/question/3932712

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