Name /or ID Question 3. (a) (SPARKING PLUGS ARE ASSETS) Assume a mechanic has on

Question

Name /or ID Question 3. (a) (SPARKING PLUGS ARE ASSETS) Assume a mechanic has only two sparking plugs left in the shop. Suppose A be the event that the first sparking plug will work and B be the event that the second sparking plug will work. Further assume that both the events A and B are independent. The probability that Sparking Plugs neither of the sparking plugs will work is | 0.50 + DOC+5000) , and the probability that the second sparking-plug will work is 0.15. What is the probability that the first plug will work?

Explanation / Answer

Probability that neither of sparking plugs will work

= 0.50 + 412/ (412 + 5000)

= 0.5761

So

P(First plug or second plug work) = 1 - 0.5761 = 0.4239

Given,

P(Second plug works) = 0.15

Also since first plug working and second plug working are independent,

P(First and second plug work) = P(First plug work)*P(Second plug work)

Hence by the formula:

P(First or second plug works) = P(First plug works) + P(Second plug works) - P(First and second plug works)

0.4239 = P(First plug works) + 0.15 - 0.15*P(First plug works)

So,

P(First plug works) = (0.4239 - 0.15)/0.85

P(First plug works) = 0.3222

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