All printed circuit boards (PCB) that are manufactured at a certain plant are inspected. An analysis of the company’s record shows that 22% of all PCBs are flawed in some way. Of those that are flawed, 84% are reparable but the rest must be discarded. If a newly produced PCB is randomly selected, what is the probability that it does not have to be discarded?
the answer is .0352 i got .9648 which is 1 -.0352 so why is it the first rather than the latter?
Well, it seems like you're trying to calculate the probability that a newly produced PCB does not have to be discarded. Let's break down the problem and try to figure out where the confusion lies.
First, we know that 22% of all PCBs are flawed. This means that the probability of selecting a flawed PCB is 0.22.
Next, we are told that of those flawed PCBs, 84% are reparable while the rest must be discarded. So, out of the flawed PCBs, only 16% need to be discarded. This means that the probability of selecting a flawed PCB that needs to be discarded is 0.16.
Now, let's calculate the probability that a newly produced PCB does not have to be discarded:
Probability of selecting a non-flawed PCB = 1 - Probability of selecting a flawed PCB that needs to be discarded.
Probability of selecting a non-flawed PCB = 1 - 0.16 = 0.84.
So, the correct answer is indeed 0.84 or 84%, not 0.0352.
It seems there might have been a confusion in the calculation or a misunderstanding of the problem. Remember, probabilities should always add up to 1, so if you subtract the probability of discarding a flawed PCB from 1, you will get the probability of selecting a non-flawed PCB.
To calculate the probability that a newly produced PCB does not have to be discarded, we need to consider the probability of it being flawed and reparable, and the probability of it being flawed but not reparable.
Given that 22% of all PCBs are flawed, the probability of a newly produced PCB being flawed is 0.22.
Of those flawed PCBs, 84% are reparable. So, the probability of a flawed PCB being reparable is 0.84.
To find the probability of a newly produced PCB being flawed but not reparable, we subtract the probability of it being flawed and reparable from the probability of it being flawed:
Probability of a newly produced PCB being flawed but not reparable = 0.22 - 0.22 * 0.84 = 0.22 - 0.1848 = 0.0352.
Therefore, the correct answer is 0.0352, which represents the probability that a newly produced PCB does not have to be discarded.
To solve this problem, we can use the concept of conditional probability. Let's break down the steps to find the correct answer.
First, let's define the events:
A: The PCB is flawed.
B: The PCB can be repaired.
We are given the following probabilities:
P(A) = 0.22 (probability that a randomly selected PCB is flawed)
P(B|A) = 0.84 (probability that a flawed PCB can be repaired)
We are required to find P(B), which is the probability that a newly produced PCB does not have to be discarded.
To solve this, we can use the law of total probability, which states that the probability of an event can be computed by conditioning on different events. In this case, we condition on whether the PCB is flawed or not flawed.
P(B) can be calculated as:
P(B) = P(B|A) * P(A) + P(B|A') * P(A')
Where A' is the complement of event A, i.e., the PCB is not flawed.
To find P(B|A'), we subtract P(B|A) from 1 since P(B|A') is the probability of PCBs being discarded, which is the complement of P(B|A).
Let's calculate it step by step:
P(B|A') = 1 - P(B|A) = 1 - 0.84 = 0.16
Now, we can substitute these values into the formula:
P(B) = P(B|A) * P(A) + P(B|A') * P(A')
= 0.84 * 0.22 + 0.16 * (1 - 0.22)
= 0.1848 + 0.128
= 0.3128
Therefore, the probability that a newly produced PCB does not have to be discarded is 0.3128, not 0.0352.
It seems like there may have been a calculation error or misunderstanding in your previous calculation. To double-check, you can redo the calculations using the correct values to ensure accuracy.