A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and 2 torches are defective. If a person takes out two items

Gmat_mission wrote: ↑

Thu Dec 17, 2020 12:44 pm

A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and 2 torches are defective. If a person takes out two items from the bag at random, what is the probability that both are bulbs or both are in working condition?

A. 8/35
B. 3/7
C. 23/35
D. 6/7
E. None of these

Answer: C

Solution:

If we let B = both items are bulbs and W = both items are in working condition, then:

Pr(B or W) = Pr(B) + Pr(W) - Pr(B and W)

Let’s first determine Pr(B), the probability that both items are bulbs. Since there are 10 bulbs and we must select 2, the number of ways to select the 2 bulbs is 10C2 = 10!/[2!(10-2)!] = (10 x 9)/2! = 5 x 9 = 45. The total number of ways to select 2 items from the group of 15 items is 15C2 = 15!/[2!(15-2)!] = (15 x 14)/2! = 15 x 7 = 105. Thus, the probability of selecting 2 bulbs is 45/105 = 9/21 = 3/7.

Next let’s determine Pr(W), the probability that both items are in working condition. Since there are a total of 15 items and 5 are defective, we have 15 - 5 = 10 items in working condition.

Thus, the number of ways to select 2 working items is 10C2 = 10!/[2!(10-2)!] = (10 x 9)/2! = 45. The total number of ways to select 2 items from the group of 15 items is 15C2 = 15!/[2!(15-2)!] = (15 x 14)/2! = 15 x 7 = 105. Thus, the probability of selecting 2 working items is 45/105 = 9/21 = 3/7.

Finally, let’s determine Pr(B and W), the probability of selecting 2 working bulbs. Since there are 7 working bulbs and we must select 2 of them, the number of ways to select those 2 bulbs is 7C2 = 7!/[2!(7-2)!] = (7 x 6)/2! = 21. Since the total number of outcomes is 105, the probability of selecting 2 working bulbs is 21/105 = 1/5.

Thus, the probability of selecting 2 bulbs or 2 working items is:

3/7 + 3/7 - 1/5 = 6/7 - 1/5 = 30/35 - 7/35 = 23/35

Answer: C