Question

An executive committee consisted of 10 members: 4 women and 6 men. Three members were selected at random to be sent to a meeting in Hawaii. A blindfolded woman drew 3 of the 10 names from a hat to determine who would go. All 3 names drawn were women's. What was the probability of such luck? (please enter your answer in decimal form)

Answer 1

Solution:

There are total 10 members:

4 Women and 6 men

The probability that first selected ‘member is women’ is given as below:

Probability that first member is women = 4/10 = 0.4

Probability that second member is women = 3/9 = 0.333333

(Because there would be only three women will be available because one woman is already selected at first draw.)

In such a way,

Probability that third member is women = 2/8 = 0.25

So, by using multiplication rule of probability, we have

Probability that all three names drawn are women = 0.4*0.333333*0.25 = 0.033333

Required probability = 0.033333