SOLUTION: Sherman is repairing his car. He has removed the six spark plugs. four are good and two are defective. He now selects one plug and then, without replacing it selects a second plug.

Question 409442: Sherman is repairing his car. He has removed the six spark plugs. four are good and two are defective. He now selects one plug and then, without replacing it selects a second plug.
1. what is the probability that both spark plugs selected are good?
2. How is the [a] possible outcome and [b] favourable outcome calculated?
3. what is the probability that both spark plugs selected are good if we know that the first plug is good?
4. what is the probability of defective plugs?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Sherman is repairing his car. He has removed the six spark plugs. four are good and two are defective. He now selects one plug and then, without replacing it selects a second plug.
P(good) = 4/6 = 2/3
P(bad) = 2/6 = 1/3
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1. what is the probability that both spark plugs selected are good?
P(2 good) = P(good)*P(good|good) = (4/6)*(3/5) = 12/30 = 2/5
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2. How is the
[a] possible outcome
[b] favourable outcome calculated?
---
Not sure what you mean by that.
===================================
3. what is the probability that both spark plugs selected are good if we know that the first plug is good?
P(good|good) = 3/5
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4. what is the probability of defective plugs?
How many?
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Cheers,
Stan H.
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