A box contains 12 bulbs with 3 defective ones. If two bulbs are drawn from the box, what is the probability that both bulbs are defective. both are non defective and one bulb is defective?
prob(both defective) = (3/12)(2/11) = 6/132
prob(both non-defective) = (9/12)(8/11) = 72/132
prob(one defective) = (3/12)(9/11) + (9/12)(3/11) = 54/132
(did you notice that 6/132 + 72/132 + 54/132 = 1 ? )
To find the probability of drawing bulbs with specific characteristics, we need to use combinations and the concept of probability.
Let's break down the problem step by step:
Step 1: Find the total number of ways to select 2 bulbs out of the 12.
This can be calculated using combinations. The formula for combinations is:
C(n, r) = n! / (r!(n-r)!)
In this case, we have 12 bulbs (n) and we want to select 2 bulbs (r), so the formula becomes:
C(12, 2) = 12! / (2!(12-2)!) = 66
Therefore, there are 66 possible ways to select 2 bulbs out of the 12.
Step 2: Find the probability of both bulbs being defective.
Since there are 3 defective bulbs in the box, the probability of selecting a defective bulb on the first draw is 3/12. After the first defective bulb is drawn, there are 2 defective bulbs remaining out of the 11 remaining bulbs in the box. So, the probability of drawing a second defective bulb is 2/11.
To find the probability of both events occurring, we multiply the probabilities:
P(both defective) = (3/12) * (2/11) = 6/132 = 1/22
Therefore, the probability of both bulbs being defective is 1/22.
Step 3: Find the probability of both bulbs being non-defective.
There are 12 total bulbs, and 3 of them are defective. So, the remaining 9 bulbs are non-defective. The probability of drawing a non-defective bulb on the first draw is 9/12. After the first non-defective bulb is drawn, there are 8 non-defective bulbs remaining out of the 11 remaining bulbs in the box. So, the probability of drawing a second non-defective bulb is 8/11.
To find the probability of both events occurring, we multiply the probabilities:
P(both non-defective) = (9/12) * (8/11) = 72/132 = 18/33 = 6/11
Therefore, the probability of both bulbs being non-defective is 6/11.
Step 4: Find the probability of one bulb being defective and one being non-defective.
For this case, we have two possibilities: defective-non-defective or non-defective-defective.
First, let's consider drawing a defective bulb followed by a non-defective bulb:
The probability of drawing a defective bulb on the first draw is 3/12. After the first defective bulb is drawn, there are 9 non-defective bulbs remaining out of the 11 remaining bulbs in the box. So, the probability of drawing a non-defective bulb on the second draw is 9/11.
P(defective-non-defective) = (3/12) * (9/11) = 27/132 = 9/44
Now, let's consider drawing a non-defective bulb followed by a defective bulb:
The probability of drawing a non-defective bulb on the first draw is 9/12. After the first non-defective bulb is drawn, there are still 3 defective bulbs remaining out of the 11 remaining bulbs in the box. So, the probability of drawing a defective bulb on the second draw is 3/11.
P(non-defective-defective) = (9/12) * (3/11) = 27/132 = 9/44
To find the probability of either event occurring, we add the probabilities:
P(one defective and one non-defective) = P(defective-non-defective) + P(non-defective-defective) = 9/44 + 9/44 = 18/44 = 9/22
Therefore, the probability of one bulb being defective and one being non-defective is 9/22.
In summary:
- The probability of both bulbs being defective is 1/22.
- The probability of both bulbs being non-defective is 6/11.
- The probability of one bulb being defective and one being non-defective is 9/22.