5. These two questions are problems that pertain to sample spaces with equally (
ID: 3315942 • Letter: 5
Question
5. These two questions are problems that pertain to sample spaces with equally (a) (10 points) A factory produces fuses, which are packaged in boxes of 10. Three fuses are selected at random from each box for inspection. The box is rejected if at least one of these three fuses is defective. What is the probability that a box containing five defective fuses will be rejected? (b) (10 points) Three people are chosen at random. What is the probability that at least two of them were born on the same day of the week?Explanation / Answer
Solution:
5) (a) It is not impossible to answer, assuming every fuse has the same likelihood of being selected.
A box containing 5 defective fuses will contain 5 non-defective fuses. The probability of the box being rejected = the probability that the 3 fuses chosen include at least 1 of the defective - (probability that the 3 fuses chosen are all from the 8 non-defective ones)
= 1 - (number of ways of choosing 3 fuses out of 8)/(number of ways of choosing 3 fuses out of 14)
= 1 - (5C3)/(10C3)
= 1 - (5 * 4 * 3)/(10 * 9 * 8)
= 1 - 360/720
= 0.5
(b) Chance of at least two were born on the same day of the week = 1 - Chance of all three born on different days of the week
Chance of three random people born on different days of the week = 1 * 6/7 * 5/7 = 30/49
Chance of at least two were born on the same day of the week = 1 - 30/49 = 19/49
5) (a) It is not impossible to answer, assuming every fuse has the same likelihood of being selected.
A box containing 5 defective fuses will contain 5 non-defective fuses. The probability of the box being rejected = the probability that the 3 fuses chosen include at least 1 of the defective - (probability that the 3 fuses chosen are all from the 8 non-defective ones)
= 1 - (number of ways of choosing 3 fuses out of 8)/(number of ways of choosing 3 fuses out of 14)
= 1 - (5C3)/(10C3)
= 1 - (5 * 4 * 3)/(10 * 9 * 8)
= 1 - 360/720
= 0.5
(b) Chance of at least two were born on the same day of the week = 1 - Chance of all three born on different days of the week
Chance of three random people born on different days of the week = 1 * 6/7 * 5/7 = 30/49
Chance of at least two were born on the same day of the week = 1 - 30/49 = 19/49
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