A quality-control program at a plastic bottle production line involves
inspecting finished bottles for flaws such as microscopic holes. The proportion of bottles that actually
have such a flaw is only 0.0002. If a bottle has a flaw, the probability is 0.995 that it will fail the
inspection. If a bottle does not have a flaw, the probability is 0.99 that it will pass the inspection (use
Bayes’s Rule).
a. If a bottle fails inspection, what is the probability that it has a flaw?
b. Which of the following is the more correct interpretation of the answer to part(a)?
i. Most bottles that fail inspection do not have a flaw.
ii. Most bottles that pass inspection do have a flaw.
c. If a bottle passes inspection, what is the probability that it does not have a flaw?
d. Which of the following is the more correct interpretation of the answer to part(c)?
i. Most bottles that fail inspection do have a flaw.
ii. Most bottles that pass inspection do not have a flaw.
e. Explain why a small probability in part (a) is not a problem, so long as the probability in part (c)
is large.
a. P(A/B)=0.01952
b. i. most of the bottles that fail inspection do not have a flow
c. P(A'/B')=0.999998
d.Most bottles that pass inspection do not have a flaw.
e.
a. If a bottle fails inspection, the probability that it has a flaw can be calculated using Bayes's Rule. Let's denote F as the event that the bottle has a flaw and I as the event that the bottle fails inspection. The probability of F given I, denoted P(F|I), can be calculated as follows:
P(F|I) = (P(I|F) * P(F)) / P(I)
P(I|F) is the probability of failing inspection given that the bottle has a flaw, which is given as 0.995. P(F) is the proportion of bottles that actually have a flaw, which is given as 0.0002. P(I) is the probability of failing inspection, which can be calculated by considering both the cases where the bottle has a flaw and doesn't have a flaw:
P(I) = P(I|F) * P(F) + P(I|not F) * P(not F)
P(I|not F) is the probability of failing inspection given that the bottle does not have a flaw, which is the complement of passing inspection, given as 1 - 0.99 = 0.01. P(not F) is the complement of P(F), which is 1 - 0.0002.
Using these values, we can calculate P(I):
P(I) = (0.995 * 0.0002) + (0.01 * (1 - 0.0002))
Now we can substitute these values into Bayes's Rule to find P(F|I):
P(F|I) = (0.995 * 0.0002) / ((0.995 * 0.0002) + (0.01 * (1 - 0.0002)))
b. The more correct interpretation of the answer to part (a) is i. Most bottles that fail inspection do not have a flaw. This is because the proportion of bottles that actually have a flaw is very small (0.0002), while the probability of failing inspection given that the bottle does not have a flaw is relatively large (0.01).
c. If a bottle passes inspection, the probability that it does not have a flaw can also be calculated using Bayes's Rule. Let's denote not F as the event that the bottle does not have a flaw and P as the event that the bottle passes inspection. The probability of not F given P, denoted P(not F|P), can be calculated similarly to part (a) as follows:
P(not F|P) = (P(P|not F) * P(not F)) / P(P)
P(P|not F) is the probability of passing inspection given that the bottle does not have a flaw, which is given as 0.99. P(not F) is the complement of P(F), which is 1 - 0.0002. P(P) is the probability of passing inspection, which can be calculated by considering both the cases where the bottle has a flaw and doesn't have a flaw:
P(P) = P(P|F) * P(F) + P(P|not F) * P(not F)
P(P|F) is the probability of passing inspection given that the bottle has a flaw, which is the complement of failing inspection, given as 1 - 0.995 = 0.005.
Using these values, we can calculate P(P) and substitute them into Bayes's Rule to find P(not F|P).
d. The more correct interpretation of the answer to part (c) is ii. Most bottles that pass inspection do not have a flaw. This is because the proportion of bottles that actually have a flaw is very small (0.0002), while the probability of passing inspection given that the bottle has a flaw is relatively small (0.005).
e. The small probability in part (a) is not a problem as long as the probability in part (c) is large because the goal of the quality-control program is to detect flaws accurately. It is more important to have a high probability of correctly identifying a bottle without a flaw (part c) than correctly identifying a bottle with a flaw (part a). Having a small probability in part (a) simply means that the proportion of bottles that actually have a flaw is very low, which is a good thing from a quality-control perspective. As long as the probability in part (c) is high, indicating a high chance of correctly passing bottles without a flaw, the quality-control program can be considered effective.
a. To find the probability that a bottle has a flaw given that it fails inspection, we can use Bayes's Rule:
P(Flaw|Fail) = (P(Fail|Flaw) * P(Flaw)) / P(Fail)
P(Fail|Flaw) = 0.995 (probability of failing inspection given that the bottle has a flaw)
P(Flaw) = 0.0002 (proportion of bottles that actually have a flaw)
P(Fail) = P(Fail|Flaw) * P(Flaw) + P(Fail|No Flaw) * P(No Flaw)
= (0.995 * 0.0002) + (0.01 * 0.9998) (probability of failing inspection)
P(Flaw|Fail) = (0.995 * 0.0002) / ((0.995 * 0.0002) + (0.01 * 0.9998))
≈ 0.0198 (approximately)
Therefore, the probability that a bottle has a flaw given that it fails inspection is approximately 0.0198, or 1.98%.
b. The more correct interpretation of the answer to part (a) is i. Most bottles that fail inspection do not have a flaw. This is because the probability of a bottle having a flaw given that it fails inspection is quite low (approximately 1.98%).
c. To find the probability that a bottle does not have a flaw given that it passes inspection, we can use Bayes's Rule:
P(No Flaw|Pass) = (P(Pass|No Flaw) * P(No Flaw)) / P(Pass)
P(Pass|No Flaw) = 0.99 (probability of passing inspection given that the bottle does not have a flaw)
P(No Flaw) = 1 - P(Flaw) = 1 - 0.0002 = 0.9998 (proportion of bottles that do not have a flaw)
P(Pass) = P(Pass|No Flaw) * P(No Flaw) + P(Pass|Flaw) * P(Flaw)
= (0.99 * 0.9998) + (0.01 * 0.0002) (probability of passing inspection)
P(No Flaw|Pass) = (0.99 * 0.9998) / ((0.99 * 0.9998) + (0.01 * 0.0002))
≈ 0.9998 (approximately)
Therefore, the probability that a bottle does not have a flaw given that it passes inspection is approximately 0.9998, or 99.98%.
d. The more correct interpretation of the answer to part (c) is ii. Most bottles that pass inspection do not have a flaw. This is because the probability of a bottle not having a flaw given that it passes inspection is quite high (approximately 99.98%).
e. In this case, a small probability in part (a) is not a problem as long as the probability in part (c) is large. This is because part (a) considers the probability of having a flaw given that a bottle fails inspection, which is inherently a low probability (approximately 1.98%). On the other hand, part (c) considers the probability of not having a flaw given that a bottle passes inspection, which is a high probability (approximately 99.98%). The focus of the quality-control program is to accurately identify bottles without flaws, so a high probability of correctly identifying bottles without flaws is more important than the low probability of correctly identifying bottles with flaws.
To solve these questions using Bayes's Rule, we need to understand the given information. Let's define some variables:
Let F be the event that a bottle has a flaw.
Let P be the event that a bottle passes the inspection.
Given probabilities:
P(F) = 0.0002 (probability of a bottle having a flaw)
P(P|F) = 0.005 (probability of a bottle failing inspection given it has a flaw)
P(P|F') = 0.01 (probability of a bottle passing inspection given it does not have a flaw)
a. To find the probability that a bottle has a flaw given that it fails inspection (P(F|P)), we can use Bayes's Rule:
P(F|P) = (P(P|F) * P(F)) / P(P)
P(F|P) = (0.005 * 0.0002) / P(P)
But we don't know P(P), so let's calculate it.
P(P) = P(P|F) * P(F) + P(P|F') * P(F')
P(P) = 0.005 * 0.0002 + 0.01 * (1 - 0.0002) = 0.0000105
Substituting this value back in, we can find P(F|P):
P(F|P) = (0.005 * 0.0002) / 0.0000105 ≈ 0.095 (approximately 0.095)
Therefore, the probability that a bottle has a flaw given that it fails inspection is approximately 0.095 or 9.5%.
b. The more correct interpretation of the answer to part (a) is i. Most bottles that fail inspection do not have a flaw. This is because the probability of a bottle having a flaw given it fails inspection is relatively low (approximately 0.095 or 9.5%).
c. To find the probability that a bottle does not have a flaw given that it passes inspection (P(F'|P)), we can use Bayes's Rule:
P(F'|P) = (P(P|F') * P(F')) / P(P)
P(F'|P) = (0.01 * (1 - 0.0002)) / P(P)
Again, we don't know P(P), so let's calculate it using the previous value:
P(P) = P(P|F) * P(F) + P(P|F') * P(F')
P(P) = 0.005 * 0.0002 + 0.01 * (1 - 0.0002) = 0.0000105 (same as before)
Substituting this value back in, we can find P(F'|P):
P(F'|P) = (0.01 * (1 - 0.0002)) / 0.0000105 ≈ 0.999 (approximately 0.999)
Therefore, the probability that a bottle does not have a flaw given that it passes inspection is approximately 0.999 or 99.9%.
d. The more correct interpretation of the answer to part (c) is ii. Most bottles that pass inspection do not have a flaw. This is because the probability of a bottle not having a flaw given it passes inspection is relatively high (approximately 0.999 or 99.9%).
e. A small probability in part (a) is not a problem as long as the probability in part (c) is large because the goal is to minimize false positives (bottles without flaws being flagged) rather than false negatives (flawed bottles passing inspection). In this scenario, it's more important to ensure the bottles that pass inspection are truly free of flaws, and a high probability of not having a flaw given passing inspection (part c) indicates a strong quality-control process.