Answer:
Explanation:
Given that a manufacturing 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.
a) The probability that the first part is suitable.
= no of suitable/total =
![(15)/(20) =0.75](https://img.qammunity.org/2021/formulas/mathematics/college/kqd8d1360jm8pasx5o5yivclw5050a8ilg.png)
b) the probability that the second part is also suitable/first part was suitable
= Prob that first two parts suitable/Prob first part suitable
=
![((15C2)/(20C2) )/(0.75) \\= 0.7368](https://img.qammunity.org/2021/formulas/mathematics/college/r3gzf6frf91bul0ajejsum1j66058fw48y.png)
c) If the first part is suitable, find the probability that the second part is not suitable
= Prob I part suitable and second not suitable/Prob first part suitable
=
![((15)/(20)(5)/(19) )/(0.75) \\= 0.2632](https://img.qammunity.org/2021/formulas/mathematics/college/ymbngrf7qasa6kv48hbycwtu14sshumzx7.png)