Amanufacturing firm just received a shipment of 20 assembly parts, of slightly varied sizes, from a vendor. the manager knows that there are only 15 parts in the shipment that would be suitable. he examines these parts one at a time. a. find the probability that the first part is suitable. (round your answer to 2 decimal places.) probability b. if the first part is suitable, find the probability that the second part is also suitable. (round your answer to 4 decimal places.) probability c. if the first part is suitable, find the probability that the second part is not suitable. (round your answer to 4 decimal places.) probability

a. 0.75

b. 0.5526

c.  0.1974

Step-by-step explanation:

Probability is typically expressed as a fraction where the numerator is the number of desired outcomes and the denominator is the total number of outcomes. In this case the suitable parts is the desired outcome and the total shipment is the total parts:

The probability the both the first and second part are suitable:

Since the probability of getting a suitable part the first time is 15/20, then there is one less suitable part and one less part overall to choose on the second time, or 14/19.

Lastly, the probability of choosing a suitable part the first time is again 15/20, but the second part not being suitable would be 5/19:

n = -5

Step-by-step explanation:

Divide both sides by 3 to get x by itself