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
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.
To find the probability of at most 3 lamps being defective when choosing 2 lamps from a pack of 12, we will calculate the probability of each case where 0, 1, 2, or 3 lamps are defective, and then add them together.
Total number of lamps = 12
Number of defective lamps = 4
Case 0: No defective lamps
To choose 2 non-defective lamps from 8 (since there are 12 - 4 = 8 non-defective lamps) out of 12 in total:
Probability of case 0 = (8C2) / (12C2) = (8! / (2! * (8-2)!)) / (12! / (2! * (12-2)!)) = (8 * 7) / (12 * 11) = 56 / 132 = 7 / 33
Case 1: 1 defective lamp
To choose 1 defective lamp from 4 and 1 non-defective lamp from 8 out of 12 in total:
Probability of case 1 = (4C1) * (8C1) / (12C2) = (4! / (1! * (4-1)!)) * (8! / (1! * (8-1)!)) / (12! / (2! * (12-2)!)) = (4 * 8) / (66 * 22) = 32 / 1452 = 8 / 363
Case 2: 2 defective lamps
To choose 2 defective lamps from 4 out of 12 in total:
Probability of case 2 = (4C2) / (12C2) = (4! / (2! * (4-2)!)) / (12! / (2! * (12-2)!)) = (6) / (66 * 22) = 1 / 242
Case 3: 3 defective lamps
To choose 3 defective lamps from 4 and no non-defective lamps, which is not possible in this case:
Probability of case 3 = 0
Now, we add up the probabilities for each case:
Total probability = probability of case 0 + probability of case 1 + probability of case 2 + probability of case 3
Total probability = 7/33 + 8/363 + 1/242 + 0
Therefore, the probability of at most 3 lamps being defective when choosing 2 lamps from a pack of 12 lamps is approximately 0.25, or 25%.