A box contains 12 bulbs with 3 defective ones. If two bulbs are drawn from the box together.what is the probability that
a)both bulbs are defetive
b)both are non defective
c)one bulb is defective?
0.25
I AM NOT KNOWN
9!6/3!
To calculate the probabilities, we need to first determine the total number of possibilities and the favorable outcomes for each scenario.
a) Both bulbs are defective:
To find the probability that both bulbs are defective, we need to calculate the ratio of the number of ways to choose 2 defective bulbs out of the 3 defective bulbs to the total number of ways to choose any 2 bulbs out of the 12 available bulbs.
Total number of possibilities: Choosing 2 bulbs out of 12 is denoted as C(12, 2), which is calculated as 12! / (2! * (12-2)!) = 66.
Favorable outcomes: We want to choose 2 defective bulbs out of the 3 available, which is denoted as C(3, 2), calculated as 3! / (2! * (3-2)!) = 3.
Therefore, the probability of both bulbs being defective is 3/66, which can be simplified to 1/22.
b) Both bulbs are non-defective:
To find the probability that both bulbs are non-defective, we need to calculate the ratio of the number of ways to choose 2 non-defective bulbs out of the 9 non-defective bulbs to the total number of ways to choose any 2 bulbs out of the 12 available bulbs.
Total number of possibilities remains the same: 66.
Favorable outcomes: We want to choose 2 non-defective bulbs out of the 9 available, which is denoted as C(9, 2), calculated as 9! / (2! * (9-2)!) = 36.
Therefore, the probability of both bulbs being non-defective is 36/66, which can be simplified to 6/11.
c) One bulb is defective:
To find the probability that one bulb is defective, we need to calculate the ratio of the number of ways to choose 1 defective bulb out of the 3 defective bulbs and 1 non-defective bulb out of the 9 non-defective bulbs, to the total number of ways to choose any 2 bulbs out of the 12 available bulbs.
Total number of possibilities remains the same: 66.
Favorable outcomes: We want to choose 1 defective bulb out of the 3 available (C(3, 1) = 3), and 1 non-defective bulb out of the 9 available (C(9, 1) = 9).
Therefore, the probability of one bulb being defective is (3 * 9) / 66, which can be simplified to 1/2.