Question

Suppose that 84% of bolts and 79% of nails meet specifications. One bolt and one nail are chosen independently. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.What is the probability that at least one of them meets specifications? (Round the final answer to four decimal places.)

Answer 1

Answer:

P(At least one of them meet specifications) = 0.9664

Step-by-step explanation:

Hi,

We have the following data:

P(Bolt meets specifications) = P(B) = 0.84

P(Nut meets specifications) = P(N) = 0.79

Since there are independent events, which means each of them have no impact on the outcome of the other, we can find out the probability of each item not meeting the specifications.

We know that the total probability of any event always equals 1.

P(Bolt not meeting specifications) = P(B') = 1 - 0.84 = 0.16

P(Nut not meeting specifications) = P(N') = 1 - 0.79 = 0.21

The question asks us to consider a situation where at least one meets the specifications. Hence we shall have the following possibilities:

• Nut meeting specification, Bolt meeting specification
• Nut not meeting specification, Bolt meeting specification
• Nut meeting specification, Bolt not meeting specification

In all these case scenarios, remember the rules of probability. When two events occur together, "and" → multiplication ; when two either of the event can occurs, "or" → addition.

$P(B\ and\ N) + P (B'\ and\ N) + P (B\ and\ N')$

⇒$(0.84$ ×$0.79)$ $+ (0.84$ × $0.21) + (0.79$ × $0.16)$

$0.6636 + 0.1764 + 0.1264\\=0.9664$

I hope this solves your query.