Question 121858: A certain airplane has two independent alternators to provide electrical power. The probability that a given alternator will fail on a 1-hour flight is .02. What is the probability that
a) both will fail?
b) Neither will fail?
c) One or the other will fail? Show all steps carefully.
Could you kindly help me. This is new to me.
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! The probability of two independent events occurring is equal to the product of each of
the independent events. And the probabilities of some outcome must total to be 1.00.
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In the discussion below, P(event) means the probability of the event being discussed.
For example, P(Alt #1 Fails) means the probability that Alternator #1 fails.
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Let's apply the above rules to solving this problem. Since the probability of a single Alternator
failing is 0.02, then the probability of it not failing must be 0.98 because those are the
only two outcomes possible and the total of all the outcomes for an Alternator must be 1.0
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The same is true for each of the two Alternators. The probability of failure is 0.02 and the
probability of working OK is 0.98 for each.
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What are the possible outcomes on a one-hour flight? The four possibilities are:
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(1) Alternator #1 Fails and Alternator #2 Fails
(2) Alternator #1 Fails and Alternator #2 Works OK
(3) Alternator #1 Works OK and Alternator #2 Fails
(4) Alternator #1 Works OK and Alternator #2 Works OK
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The probability of (1) occurring is just the product of the two probabilities. So the
probability of both Alternators failing is:
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P(both fail) = P(Alt #1 Fails) * P(Alt #2 Fails) = 0.02 * 0.02 = 0.0004
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Next, let's look at the probability of (4) occurring. It also is the product of the two
probabilities. So the probability of both Alternators working OK is:
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P(both OK) = P(Alt #1 Works OK) * P(Alt #2 Works OK) = 0.98 * 0.98 = 0.9604
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Next, let's look at the probability of (2) occurring. It also is the product of the two
probabilities. So the probability of Alternator #1 failing and Alternator #2 working OK is:
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P(Alt #1 Fail and Alt #2 OK) = P(Alt #1 Fails) * P(Alt #2 Works OK) = 0.02 * 0.98 = 0.0196
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Finally, let's look at the probability of (3) occurring. It also is the product of the two
probabilities. So the probability of Alternator #1 working OK and Alternator #2 failing is:
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P(Alt #1 OK and Alt #2 Fails) = P(Alt #1 Works OK) * P(Alt #2 Fails) = 0.98 * 0.02 = 0.0196
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In either case (2) or case (3) one of the Alternators fails and the other one works OK.
The probability of that happening is the sum of the two probabilities for these two cases. So
the probability for one Alternator failing is:
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P(one fail, one OK) = P(Alt #1 Fail and Alt #2 Works OK) + P(Alt #1 Works OK and Alt #2 Fails)
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Substituting the probabilities results in:
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P(one fail, one OK) = 0.0196 + 0.0196 = 0.0392
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In summary, the three answers you are looking for are:
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(a) probability both will fail = 0.0004
(b) probability neither will fail (both will work OK) = 0.9604
(c) probability one will fail = 0.0392
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The probability of some outcome (all four of the combinations we said are the only things
that can happen) must total to be 1.0 because something must happen. So we can check our
answers by adding them up and see if they total 1.0. By adding 0.0004, 0.9604, and 0.0392
you do get an answer of 1.0 ... so we are correct.
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Hope this helps you to understand the problem a little better.
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