Question

CasL ON 15 Bldudal There are 3 economists, 4 engineers, 2 statisticians, and 1 doctor. A committee of 4 among them is to be formed. Find the probability that the committee (i) consists of one of each kind, (ii) has at least one economist, (iii) has the doctor as a member and three others being selected in for

a. (i) 0.025, (ii) 0.875, (iii) 0.067
b. (i) 0.067, (ii) 0.875, (iii) 0.025
c. (i) 0.025, (ii) 0.067, (iii) 0.875
d. (i) 0.067, (ii) 0.025, (iii) 0.875

Answer 1

The probability that the committee consists of one of each kind is 0.114, the probability of having at least one economist is 0.833, and the probability of the doctor being a member with three others selected is 0.4.

Step-by-step explanation:

To find the probability of forming a committee with specific characteristics, we can use the concept of combinations. The total number of ways to select 4 people from 10 is the combination of 10 choose 4, which is calculated as:

C(10, 4) = 10! / (4! * (10-4)!) = 210

(i) For the first part, we need to select one economist, one engineer, one statistician, and the doctor. The number of ways to do this is: C(3, 1) * C(4, 1) * C(2, 1) * C(1, 1) = 3 * 4 * 2 * 1 = 24. Therefore, the probability is given by 24 / 210 = 0.114.

(ii) To find the probability of having at least one economist, we can calculate the probability of having no economist and subtract it from 1. The number of ways to form a committee with no economist is: C(7, 4) = 35. Therefore, the probability is 1 - 35 / 210 = 0.833.

(iii) For the third part, we need to select the doctor and three more people. The number of ways to do this is: C(1, 1) * C(9, 3) = 1 * 84 = 84. Therefore, the probability is 84 / 210 = 0.4.