A box contain 12 cell phones of which 4 are defective.all phones look alike.3 phones are drawn randomly what is the probability.1 all the 3 bulbs are defective.2 at least 2 of the bulbs chosen are defective.3 at most 2 of the bulbs chosen are defective.hint use combination analysis
prob(defective) = 4/12 = 1/3
prob(not defective) = 2/3
1. all 3 defective ---- prob = (1/3)^3 = 1/27
2. at least 2 defective
----> 2 defective or 3 defective
prob = C(3,2)(1/3)^2 (2/3) + (1/3)^3
= 3(2/27) + 1/27 = 7/27
3. at most 2 ---> 0 defective, 1 defect, or 2 defective
or exclude all 3 defectiv
= 1 - 1/27 = 26/27
BTW,
Where do the bulbs enter the picture when talking about cell phones
To find the probabilities, we need to use combination analysis and the concept of probability.
Combination analysis involves calculating the number of ways to choose a certain number of objects from a larger set. In this case, we will be using combinations because the order in which the phones are drawn doesn't matter.
1. Probability that all 3 phones are defective:
Out of the 12 phones, 4 are defective. To calculate the probability that all 3 phones drawn are defective, we need to calculate the number of ways to choose 3 phones from the 4 defective phones and divide it by the total number of ways to choose 3 phones from the 12 phones in the box.
Number of ways to choose 3 phones from the 4 defective phones = C(4, 3) = 4 (since there are only 4 defective phones)
Total number of ways to choose 3 phones from the 12 phones = C(12, 3) = 220 (using the combination formula C(n, r) = n! / (r!(n-r)!)
So, the probability that all 3 phones drawn are defective is 4/220 = 1/55.
2. Probability that at least 2 of the phones chosen are defective:
To calculate this probability, we need to find the number of ways to choose 2 defective phones from the 4 defective phones and multiply it by the number of ways to choose 1 working phone from the 8 working phones in the box. Then, we divide it by the total number of ways to choose 3 phones from the 12 phones in the box.
Number of ways to choose 2 defective phones from the 4 defective phones = C(4, 2) = 6
Number of ways to choose 1 working phone from the 8 working phones = C(8, 1) = 8
Total number of ways to choose 3 phones from the 12 phones = C(12, 3) = 220
So, the probability that at least 2 of the phones chosen are defective is (6 * 8) / 220 = 48/220 = 6/55.
3. Probability that at most 2 of the phones chosen are defective:
To calculate this probability, we need to find the probability of the cases in which 0, 1, or 2 of the phones chosen are defective and add them together.
The probability of choosing 0 defective phones is the probability that all 3 chosen phones are non-defective. We can calculate it by finding the number of ways to choose 3 non-defective phones from the 8 non-defective phones and dividing it by the total number of ways to choose 3 phones from the 12 phones.
Number of ways to choose 3 non-defective phones from the 8 non-defective phones = C(8, 3) = 56
So, the probability of choosing 0 defective phones is 56/220 = 14/55.
The probability of choosing 1 defective phone is the probability that exactly 1 of the phones chosen is defective. We can calculate it by finding the number of ways to choose 1 defective phone from the 4 defective phones and multiplying it by the number of ways to choose 2 non-defective phones from the 8 non-defective phones. Then, we divide it by the total number of ways to choose 3 phones from the 12 phones.
Number of ways to choose 1 defective phone from the 4 defective phones = C(4, 1) = 4
Number of ways to choose 2 non-defective phones from the 8 non-defective phones = C(8, 2) = 28
So, the probability of choosing 1 defective phone is (4 * 28) / 220 = 4/55.
Finally, the probability of choosing 2 defective phones can be calculated similarly.
So, the probability that at most 2 of the phones chosen are defective is (14 + 4 + probability of choosing 2 defective phones)/220 = (14 + 4 + probability of choosing 2 defective phones)/55.