smartphone repair company received 10 dysfunctional iPhones and 12 dysfunc-
tional Samsung Galaxy phones. Assuming that the company will repair only 6 devices that
are selected randomly, calculate the following probabilities
a) 3 iPhones and 3 Samsung Galaxy phones will be repaired.
b) At most 4 Samsung Galaxy phones will be repaired.
1
a) To calculate the probability that 3 iPhones and 3 Samsung Galaxy phones will be repaired, we need to use the hypergeometric distribution.
The total number of dysfunctional iPhones is 10, and the total number of dysfunctional Samsung Galaxy phones is 12. The total number of dysfunctional phones is 22.
We want to calculate the probability of selecting 3 iPhones and 3 Samsung Galaxy phones, so we choose 3 from the 10 iPhones and 3 from the 12 Samsung Galaxy phones.
The probability is given by the formula:
P(3 iPhones and 3 Samsung Galaxy phones) = (10 choose 3) * (12 choose 3) / (22 choose 6)
Using combinatorial notation, this can be expressed as:
P(3 iPhones and 3 Samsung Galaxy phones) = (10C3) * (12C3) / (22C6)
Calculating this will give us the probability.
b) To calculate the probability that at most 4 Samsung Galaxy phones will be repaired, we need to add up the probabilities of selecting 0, 1, 2, 3, and 4 Samsung Galaxy phones that will be repaired.
For each case, we use the hypergeometric distribution to calculate the probability.
The probability of selecting 0 Samsung Galaxy phones that will be repaired is given by:
P(0 Samsung Galaxy phones repaired) = (10 choose 6) * (12 choose 0) / (22 choose 6)
Similarly, the probabilities for selecting 1, 2, 3, and 4 Samsung Galaxy phones that will be repaired can be calculated using the hypergeometric distribution.
Finally, we add up all these probabilities to get the probability of having at most 4 Samsung Galaxy phones repaired.
To calculate the probabilities, we need to use the concept of combinations.
a) To calculate the probability of repairing 3 iPhones and 3 Samsung Galaxy phones:
Step 1: Find the total number of ways to select 6 phones out of the given iPhones and Samsung Galaxy phones.
Total ways = C(22, 6) = 22! / (6!(22-6)!) = 134,596
Step 2: Find the number of ways to select 3 iPhones out of 10 and 3 Samsung Galaxy phones out of 12.
Number of ways = (C(10, 3)) * (C(12, 3)) = (10! / (3!(10-3)!)) * (12! / (3!(12-3)!)) = 120 * 220 = 26,400
Step 3: Calculate the probability:
Probability = Number of favorable outcomes / Total number of outcomes
Probability = 26,400 / 134,596 ≈ 0.196 or 19.6%
b) To calculate the probability of at most 4 Samsung Galaxy phones being repaired:
Step 1: Find the total number of ways to select 6 phones out of the given iPhones and Samsung Galaxy phones.
Total ways = C(22, 6) = 134,596 (same as before)
Step 2: Find the number of ways to select 0, 1, 2, 3, or 4 Samsung Galaxy phones.
Number of ways = (C(10, 6)) + (C(10, 5)) + (C(10, 4)) + (C(10, 3)) + (C(10, 2)) = 210 + 252 + 210 + 120 + 45 = 837
Step 3: Calculate the probability:
Probability = Number of favorable outcomes / Total number of outcomes
Probability = 837 / 134,596 ≈ 0.006 or 0.6%
To calculate the probabilities, we need to use the concept of combinations.
a) To calculate the probability of repairing exactly 3 iPhones and 3 Samsung Galaxy phones, we need to calculate the number of ways we can select 3 iPhones out of 10 and 3 Samsung Galaxy phones out of 12.
The probability can be calculated as follows:
P(3 iPhones and 3 Samsung Galaxy phones) = (number of ways to select 3 iPhones out of 10) * (number of ways to select 3 Samsung Galaxy phones out of 12) / (total number of ways to select 6 phones out of 22)
The number of ways to select 3 iPhones out of 10 can be calculated using the combination formula: C(10, 3) = 10! / (3! * (10-3)!) = 120
Similarly, the number of ways to select 3 Samsung Galaxy phones out of 12 can be calculated as: C(12, 3) = 12! / (3! * (12-3)!) = 220
The total number of ways to select 6 phones out of 22 can be calculated as: C(22, 6) = 22! / (6! * (22-6)!) = 134,596
Substituting these values into the probability calculation:
P(3 iPhones and 3 Samsung Galaxy phones) = (120 * 220) / 134,596 = 0.1963
So, the probability of repairing exactly 3 iPhones and 3 Samsung Galaxy phones is approximately 0.1963.
b) To calculate the probability of at most 4 Samsung Galaxy phones being repaired, we need to calculate the probabilities of repairing 0, 1, 2, 3, and 4 Samsung Galaxy phones, and then sum up these probabilities.
The probability of repairing 0 Samsung Galaxy phones (all iPhones being repaired) can be calculated as follows:
P(0 Samsung Galaxy phones) = (number of ways to select 6 iPhones out of 10) / (total number of ways to select 6 phones out of 22)
= C(10, 6) / C(22, 6)
= 210 / 134,596
≈ 0.0016
Similarly, we can calculate the probabilities of repairing 1, 2, 3, and 4 Samsung Galaxy phones as follows:
P(1 Samsung Galaxy phone) ≈ 0.0494
P(2 Samsung Galaxy phones) ≈ 0.2332
P(3 Samsung Galaxy phones) ≈ 0.3696
P(4 Samsung Galaxy phones) ≈ 0.3155
The probability of at most 4 Samsung Galaxy phones being repaired is the sum of these probabilities:
P(at most 4 Samsung Galaxy phones) ≈ 0.0016 + 0.0494 + 0.2332 + 0.3696 + 0.3155 ≈ 0.9693
So, the probability of at most 4 Samsung Galaxy phones being repaired is approximately 0.9693.