Question

There are 7 women and 5 men in a department. A committee of 5 is to be chosen. A. How many ways can this comittee be chosen? B. How many ways can this comitee be chosen if there must be 2 women and 3 men? C. If the comitee is chosen at random, what is the probability that it will consist of 2 women and 3 men?

Answer 1

There are total 7 women and 5 men .

Total No. of people in the comittee = 5.

No. of ways to choose r object out of n different objects where order doesn't matter = C(n,r) = (n!) / [(n-r)! r!]

here C is the combination operator .

(A.)

So To calculate Total no. of ways in which committee can be chosen =  No. of ways to select 5 people from the   total 12 people including women and men = C(12,5) = 792

(B.) If there must be 2 women and 3 men in choosing the committee , we need to calculate no. of ways to chose 2 women out of 7 women and 3 men out 5 men available .

    Fundamental Principle of counting states that : if there are 'n' no. of ways to do an action and 'm' no. of ways to do another action, then no. of ways to do both action together = m * n

So No. of ways to select 2 women and 3 men = (No. of ways to select 2 women out of 7) * (No. of ways to select    3 men out of 5 )

  = C(7,2) C(5,3)= 21 * 10 = 210

(C.) IN general , Probability = No. of favourable outcomes / Total no. of outcomes

Probability of committee having 2 women and 3 men = (No. of ways to select 2 women and 3 men) / ( Total no. of ways to select a committee)

Required probability = 210 / 792 = 35 / 132 = 0.265

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