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?

To solve these probability problems, we need to understand the concept of combinations and use the formula for finding probabilities.

First, let's define some terms:
- Total number of bulbs: 12
- Number of defective bulbs: 3
- Number of non-defective bulbs: 12 - 3 = 9

a) Probability that both bulbs are defective:
To find the probability of two independent events happening together, we multiply their individual probabilities. In this case, the probability of drawing a defective bulb on the first draw is 3/12 (since there are 3 defective bulbs out of 12 total). After drawing one defective bulb, we are left with 2 defective bulbs out of 11 total for the second draw, so the probability is 2/11.
Therefore, the probability of drawing two defective bulbs is (3/12) * (2/11) = 6/132 = 1/22.

b) Probability that both bulbs are non-defective:
Similarly, the probability of drawing a non-defective bulb on the first draw is 9/12 (since there are 9 non-defective bulbs out of 12 total). After drawing one non-defective bulb, we are left with 8 non-defective bulbs out of 11 total for the second draw, so the probability is 8/11.
Therefore, the probability of drawing two non-defective bulbs is (9/12) * (8/11) = 72/132 = 6/11.

c) Probability that one bulb is defective:
To find this probability, we will consider two scenarios: drawing a defective bulb first and then a non-defective bulb, OR drawing a non-defective bulb first and then a defective bulb.

Scenario 1: Drawing a defective bulb first (probability = 3/12), followed by a non-defective bulb (probability = 9/11).
Scenario 2: Drawing a non-defective bulb first (probability = 9/12), followed by a defective bulb (probability = 3/11).

We add up the probabilities of these two scenarios to find the total probability of drawing one defective bulb.
Total probability = (3/12) * (9/11) + (9/12) * (3/11) = 27/132 + 27/132 = 54/132 = 9/22.

To summarize:
a) Probability of both bulbs being defective = 1/22
b) Probability of both bulbs being non-defective = 6/11
c) Probability of one bulb being defective = 9/22

3/20

what will be the answer

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

c) 3/12 * 9/(12-1) = ?

Use the similar process for a and b.