A manufacturing firm just received a shipment of 18 assembly parts, of slightly varied sizes, from a vendor. The manager knows that there are only 16 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.) b. If the first part is suitable, find the probability that the second part is also suitable. (Round your answer to 4 decimal places.) c. If the first part is suitable, find the probability that the second part is not suitable. (Round your answer to 4 decimal places.)

(a) 0.89

(b) 0.8823

(c) 0.1176

Step-by-step explanation:

According to the scenario, calculation of the given data are as follows,

Total number of parts = 18

Total number of suitable parts = 16

(a). Probability for the first part is suitable are as follows,

Here, favourable outcomes = 16

So, P1 = favourable outcomes ÷ Total number of parts

= 16 ÷ 18

= 0.89

(b). Probability for the second part is suitable are as follows,

Here, favourable outcomes = 16 - 1 = 15

total number of parts = 18 - 1 = 17

So, P2 = favourable outcomes ÷ Total number of parts

= 15 ÷ 17

= 0.8823

(c). Probability for the second part is not suitable are as follows,

Here, favourable outcomes = 17 - 15 = 2

Total number of parts = 17

So, P2 = favourable outcomes ÷ Total number of parts

= 2 ÷ 17

= 0.1176

idk

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