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  • Confused! kindly explain, From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is :

 From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman.  Then the probability for these committees to have more women than men, is :

 

  • Option 1)   (21/220)
  • Option 2) (3/11)
  • Option 3) (1/11)
  • Option 4) (2/23)
 

Answers (1)

best_answer

As we learnt in

Conditional Probability -

 

P\left ( \frac{A}{B} \right )= \frac{P\left ( A\cap B \right )}{P\left ( B \right )}

and

P\left ( \frac{B}{A} \right )= \frac{P\left ( A\cap B \right )}{P\left ( A \right )}

 

- wherein

where P\left ( \frac{A}{B} \right ) probability of A when B already happened.

 

 

10M and 5 W

4 member committees

P (A) = P (atleast and woman)=1-P(no woman)

=1-\frac{^{10}C_{4}}{^{15}C_{4}}

=1-\frac{210}{1365}=1-\frac{42}{273}

=1-\frac{2}{13}=\frac{11}{13}

P (more women than men) = P (B)

\frac{^{5}C_{3}\times ^{10}C_{1}+ ^{5}C_{4}\times ^{10}C_{0}}{^{15}C_{4}}

=\frac{100+5}{1365}=\frac{21}{273}=\frac{1}{13}

P\left(\frac{B}{A} \right ) =\frac{P(B\cap A)}{P(A)}=\frac{\frac{1}{13}}{\frac{11}{13}}=\frac{1}{11}

 


Option 1)

\frac{21}{220}

This option is incorrect

Option 2)

\frac{3}{11}

This option is incorrrect

Option 3)

\frac{1}{11}

This option is correct

Option 4)

\frac{2}{23}

This option is incorrrect

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Plabita

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