two lamps are to be choosen from apacK of 12 lamps where 4 are defective and the rest are non detective what is the probability of atmost 3 lamps are defective

1 answer

We can approach this problem using the concept of combinations.

There are a total of 12 lamps in the pack, out of which 4 are defective and 8 are non-defective.

To find the probability of at most 3 lamps being defective, we need to find the probability of selecting 0, 1, 2, or 3 defective lamps.

Case 1: Selecting 0 defective lamps
The number of ways to select 2 non-defective lamps out of 8 is given by 8C2 = 28.
The number of ways to select 2 lamps from the total 12 is given by 12C2 = 66.
Therefore, the probability of selecting 0 defective lamps is 28/66.

Case 2: Selecting 1 defective lamp
The number of ways to select 1 defective lamp out of 4 is given by 4C1 = 4.
The number of ways to select 1 non-defective lamp out of 8 is given by 8C1 = 8.
The number of ways to select 2 lamps from the total 12 is given by 12C2 = 66.
Therefore, the probability of selecting 1 defective lamp is (4 * 8) / 66 = 32/66.

Case 3: Selecting 2 defective lamps
The number of ways to select 2 defective lamps out of 4 is given by 4C2 = 6.
The number of ways to select 0 non-defective lamps out of 8 is given by 8C0 = 1.
The number of ways to select 2 lamps from the total 12 is given by 12C2 = 66.
Therefore, the probability of selecting 2 defective lamps is (6 * 1) / 66 = 6/66.

Case 4: Selecting 3 defective lamps
The number of ways to select 3 defective lamps out of 4 is given by 4C3 = 4.
The number of ways to select 0 non-defective lamps out of 8 is given by 8C0 = 1.
The number of ways to select 2 lamps from the total 12 is given by 12C2 = 66.
Therefore, the probability of selecting 3 defective lamps is (4 * 1) / 66 = 4/66.

Therefore, the probability of at most 3 lamps being defective is given by the sum of the above probabilities:

P(at most 3 defective lamps) = 28/66 + 32/66 + 6/66 + 4/66
= 70/66

Simplifying the fraction, we get:
P(at most 3 defective lamps) = 35/33

Therefore, the probability of at most 3 lamps being defective is 35/33 or approximately 1.06.