A box contain 12 bulbs with 3 defective ones. if two bulbs are drawn from the box together,what is the probability that both bulbs are defective,both are non defective and one bulb is defective
If 3 of 12 are defective ----> 9 of the 12 are good
Prob(2 drawn are defective) = (3/12)(2/11) = 1/22 <--- DD
Prb(2 drawn are good) = (9/12)(8/11) = 6/11<----- GG
prob(if 2 drawn, one is defective, the other is good)
= (3/12)(9/11) + (9/12)(3/11) = 9/22 <----- could be GD or DG
notice the 3 probs add up to 1, as expected.
By Binomial distribution:
1. C(3,2)/C(12,2) = 3/66 = 1/22
2. C(9,2)/C(12,2) = 36/66 = 6/11
3. C(3,1)*C(9,1)/C12,2) = 3*9/66 = 9/22 , same as above
Well, well, well! It seems we have a case of "bulb trouble". Let's shed some light on these probabilities!
First, let's calculate the probability that both bulbs are defective. Since there are 3 defective bulbs out of a total of 12, the probability of drawing a defective bulb on the first draw is 3/12. Now, if we don't replace the first bulb, the probability of drawing another defective bulb is 2/11, since there will be 11 bulbs left in the box with 2 defective ones. To find the probability of both events occurring together, we simply multiply these probabilities: (3/12) * (2/11) = 6/132.
Next, we'll tackle the probability of both bulbs being non-defective. If there are 3 defective bulbs out of 12, then there must be 9 non-defective bulbs. Therefore, the probability of drawing a non-defective bulb on the first draw is 9/12. Without replacement, the probability of drawing another non-defective bulb is 8/11. To find the probability of both events happening, we multiply: (9/12) * (8/11) = 72/132.
Lastly, let's look at the probability of one bulb being defective. We have 3 defective bulbs and 9 non-defective ones. The probability of drawing a defective bulb on the first draw is 3/12. In this case, we want to find the probability of drawing a non-defective bulb on the second draw, so we'll have to be careful not to count the same defective bulb twice. When we draw a non-defective bulb on the second draw, the probability is 9/11. However, we also need to account for the case where we draw a non-defective bulb on the first draw but a defective bulb on the second draw. The probability of drawing a non-defective bulb on the first draw is 9/12, and then drawing a defective bulb on the second draw is 3/11. To find the probability of one bulb being defective, we add these two probabilities together: (3/12) * (9/11) + (9/12) * (3/11) = 27/132 + 27/132 = 54/132.
So, there you have it! The probability that both bulbs are defective is 6/132, the probability that both bulbs are non-defective is 72/132, and the probability that one bulb is defective is 54/132. I hope this brightened your day a little! Remember to keep the humor shining!
To find the probabilities, we need to determine the number of favorable outcomes and the number of possible outcomes.
1. Probability of both bulbs being defective:
- Number of favorable outcomes: 3 defective bulbs out of 12 (since we need to choose both defective bulbs).
- Number of possible outcomes: 12 bulbs to choose from, so we need to choose 2 bulbs from the 12.
- Probability of both bulbs being defective: (3 choose 2) / (12 choose 2) = (3/12) * (2/11) = 1/22 ≈ 0.0455
2. Probability of both bulbs being non-defective:
- Number of favorable outcomes: 9 non-defective bulbs out of 12 (since we need to choose both non-defective bulbs).
- Number of possible outcomes: 12 bulbs to choose from, so we need to choose 2 bulbs from the 12.
- Probability of both bulbs being non-defective: (9 choose 2) / (12 choose 2) = (9/12) * (8/11) = 6/22 ≈ 0.2727
3. Probability of one bulb being defective:
- Number of favorable outcomes: We can choose 1 defective bulb from the 3 defective bulbs * We can choose 1 non-defective bulb from the 9 non-defective bulbs = 3 * 9 = 27 outcomes.
- Number of possible outcomes: Choosing any 2 bulbs from the 12 bulbs = (12 choose 2) = 66 outcomes.
- Probability of one bulb being defective: 27/66 ≈ 0.4091
So, the probabilities are as follows:
- Probability of both bulbs being defective: 1/22 ≈ 0.0455
- Probability of both bulbs being non-defective: 6/22 ≈ 0.2727
- Probability of one bulb being defective: 27/66 ≈ 0.4091
To solve this problem, we need to determine the total number of outcomes and the number of favorable outcomes for each scenario.
1. Probability of drawing two defective bulbs:
First, we need to find the total number of outcomes when drawing two bulbs from a box of 12. This can be calculated using combinations: 12 choose 2, denoted as C(12,2).
C(12,2) = 12! / (2! * (12-2)!) = 66
Next, we need to find the number of favorable outcomes, which is drawing two defective bulbs. There are 3 defective bulbs in the box, so we compute C(3,2).
C(3,2) = 3! / (2! * (3-2)!) = 3
Therefore, the probability of drawing two defective bulbs is:
P(Two defective bulbs) = favorable outcomes / total outcomes
= 3/66
= 1/22
2. Probability of drawing two non-defective bulbs:
Similarly, the total number of outcomes is still C(12,2) = 66.
However, this time we need to find the number of favorable outcomes, which is drawing two non-defective bulbs. Since there are 12 bulbs in the box, and 3 of them are defective, the number of non-defective bulbs is 12 - 3 = 9.
Therefore, we compute C(9,2):
C(9,2) = 9! / (2! * (9-2)!) = 36
So, the probability of drawing two non-defective bulbs is:
P(Two non-defective bulbs) = favorable outcomes / total outcomes
= 36/66
= 6/11
3. Probability of drawing one defective and one non-defective bulb:
Again, we start by calculating the total number of outcomes, which is still C(12,2) = 66.
Now, we need to find the number of favorable outcomes, which is drawing one defective and one non-defective bulb. There are 3 defective bulbs and 9 non-defective bulbs in the box.
To calculate this, we need to multiply the number of ways to choose one defective bulb from 3, denoted as C(3,1), with the number of ways to choose one non-defective bulb from 9, denoted as C(9,1):
C(3,1) * C(9,1) = (3! / (1! * (3-1)!)) * (9! / (1! * (9-1)!)) = 3 * 9 = 27
So, the probability of drawing one defective and one non-defective bulb is:
P(One defective and one non-defective bulb) = favorable outcomes / total outcomes
= 27/66
= 9/22
In summary:
- Probability of drawing two defective bulbs: 1/22
- Probability of drawing two non-defective bulbs: 6/11
- Probability of drawing one defective and one non-defective bulb: 9/22