A council consists of 10 members.four are NDC,three are NPP and three are from other parties.if a committee of three is to be selected,find the probabilities following:
I.one of each party.
II.Two NPP and one from other parties.
III.one from other parties and two NDC
To find the probabilities, we need to calculate the total number of possible outcomes and then determine the number of favorable outcomes for each situation.
In this case, we have 10 members in the council, out of which 4 are NDC, 3 are NPP, and 3 are from other parties.
I. Probability of selecting one member from each party:
Total number of possible outcomes = Total ways to select 3 members from 10 = C(10,3) = 10! / (3! * (10-3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120
Number of favorable outcomes = Number of ways to select 1 NDC, 1 NPP, and 1 member from other parties = C(4,1) * C(3,1) * C(3,1) = 4 * 3 * 3 = 36
Probability = Number of favorable outcomes / Total number of possible outcomes = 36 / 120 = 0.3
II. Probability of selecting two NPP members and one member from other parties:
Total number of possible outcomes = C(10,3) = 120 (as calculated above)
Number of favorable outcomes = Number of ways to select 2 NPP members and 1 member from other parties = C(3,2) * C(3,1) = 3 * 3 = 9
Probability = Number of favorable outcomes / Total number of possible outcomes = 9 / 120 = 0.075
III. Probability of selecting one member from other parties and two NDC members:
Total number of possible outcomes = C(10,3) = 120 (as calculated above)
Number of favorable outcomes = Number of ways to select 1 member from other parties and 2 NDC members = C(3,1) * C(4,2) = 3 * 6 = 18
Probability = Number of favorable outcomes / Total number of possible outcomes = 18 / 120 = 0.15
Therefore, the probabilities are:
I. One of each party: 0.3
II. Two NPP and one from other parties: 0.075
III. One from other parties and two NDC: 0.15