At a quality control checkpoint on a manufacturing assembly line, 8% of the items failed check A, 10% failed check B, and 2% failed both checks A and B.

a. If a product failed check A, what is the probability that it also failed check B? b. If a product failed check B, what is the probability that it also failed check A?
c. What is the probability that a product failed either check A or check B?
d. What is the probability that a product failed neither check A nor check B?

I am not good with probability and the percentages are really throwing me off. Any help would be great.

a)1/4 (25%) of those that failed A also failed B. That leaves you with 1/4 or 8% or 2% failing both.

b) Similarly, 1/5 (20%) of those that failed B also failed A.
c) Add the fractions that failed A only, B only, and both. I get 8% + 6% + 2% = 16%
That fraction failed A only, B only or both.
d) 84%

To solve this problem, we can use the concept of conditional probability along with the addition and subtraction rules of probability.

Let's define some variables:
A = event of failing check A
B = event of failing check B

a. If a product failed check A, what is the probability that it also failed check B?

We need to find P(B | A), which is read as "the probability of B given A."

Using the definition of conditional probability:
P(B | A) = P(A and B) / P(A)

We are given that 2% failed both checks A and B, so P(A and B) = 0.02.
Also, 8% failed check A, so P(A) = 0.08.

Substituting the values:
P(B | A) = 0.02 / 0.08 = 0.25

Therefore, the probability that a product failed check B given that it failed check A is 0.25.

b. If a product failed check B, what is the probability that it also failed check A?

Similarly, we need to find P(A | B), which is read as "the probability of A given B."

Using the same formula as in part (a):
P(A | B) = P(A and B) / P(B)

We already know that P(A and B) = 0.02, and given that 10% failed check B, P(B) = 0.10.

Substituting the values:
P(A | B) = 0.02 / 0.10 = 0.20

Therefore, the probability that a product failed check A given that it failed check B is 0.20.

c. What is the probability that a product failed either check A or check B?

To find this probability, we need to use the addition rule of probability.

P(A or B) = P(A) + P(B) - P(A and B)

Given that 8% failed check A (P(A) = 0.08), 10% failed check B (P(B) = 0.10), and 2% failed both checks A and B (P(A and B) = 0.02), we can substitute the values into the formula:

P(A or B) = 0.08 + 0.10 - 0.02 = 0.16

Therefore, the probability that a product failed either check A or check B is 0.16.

d. What is the probability that a product failed neither check A nor check B?

To find this probability, we subtract the probability of failing either check A or check B from 1 (since that covers all possible outcomes).

P(neither A nor B) = 1 - P(A or B)

Given that P(A or B) = 0.16 (from part c), we can substitute the value into the formula:

P(neither A nor B) = 1 - 0.16 = 0.84

Therefore, the probability that a product failed neither check A nor check B is 0.84.

I hope this explanation helps you understand how to approach and solve probability problems like this!