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July 29, 2014

July 29, 2014

Posted by **Gayla** on Wednesday, May 21, 2008 at 1:55am.

- Probability problems -
**drwls**, Wednesday, May 21, 2008 at 1:57amSome of us might be able to if you tell us the problems.

- Probability problems -
**Gayla**, Wednesday, May 21, 2008 at 2:10amThank you and here is the two problems.

1. A certain airplane has two independent alternators to provide electrical power. The probability that a certain 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.

2. The probability is 1 in 4,000,000 that a single auto trip in the United States will result in a fatality. Over a lifetime, an average U.S. driver takes 50,000 trips. (a)What is the probability of a fatal accident over a lifetime? Explain your reasoning. Hint: Assume independent events. Why might the assumption of independence be violated? (b)Why might a driver by tempted to not use a seat belt "just on this trip"?

Can you please help me with both of these problems as I am not sure how to solve them? Thank you for any and all help.

- Probability problems -
**Justina**, Saturday, May 31, 2008 at 9:26pm1.

a) 0.02*0.02= 0.04

b) 1-0.04= 0.06

c) 0.02*0.08= 0.0016

- Probability problems -

- Probability problems -
- Probability problems -
**drwls**, Wednesday, May 21, 2008 at 6:24am1. Multiply the probabilities of the separate outcopmes for each.

The probability that both will fail is (0.02)^2 = 0.0004

The probability that #1 will fail and #2 will not is 0.02*0.98 = 0.196

The probability that #2 will fail and #1 will not is also 0.01960

You need to add the last two results to get the probability that one will fail: 0.0392

The probability that neither will fail is (0.98)^2 = 0.9604

Note that the total is 0.0004 + 0.0392 + 0.9604 = 1.0000

2. The probability of surviving each trip is 1 - 4*10^-6 = 0.999996.

Take the 50,000 power of that as the probability of surviving 50,000 trips in a row with no fatality to anyone. That is 0.819. (The number seems too low - about 5% of people in the United States die in accidents; not 18%. But that is what you get using their numbers and "independent event" assumptions)

Trips are not really independent since the probability of a fatality on any trip varies with the driver's age, state of residence, time of day/night, weather conditions and the length of the trip.

(b) Some people might expect the probability of being injured in a short off-highway trip, or one in light traffic, to be less, and not want to bother with the seat belt in such a case. Statistics may prove such trips are more dangerous. (I don't know)

- Probability problems -
**drwls**, Wednesday, May 21, 2008 at 6:25am1. Multiply the probabilities of the separate outcopmes for each.

The probability that both will fail is (0.02)^2 = 0.0004

The probability that #1 will fail and #2 will not is 0.02*0.98 = 0.196

The probability that #2 will fail and #1 will not is also 0.01960

You need to add the last two results to get the probability that one will fail: 0.0392

The probability that neither will fail is (0.98)^2 = 0.9604

Note that the total is 0.0004 + 0.0392 + 0.9604 = 1.0000

2. The probability of surviving each trip is 1 - 4*10^-6 = 0.999996.

Take the 50,000 power of that as the probability of surviving 50,000 trips in a row with no fatality to anyone. That is 0.819. (The number seems too low - about 5% of people in the United States die in accidents; not 18%. But that is what you get using their numbers and "independent event" assumptions)

Trips are not really independent since the probability of a fatality on any trip varies with the driver's age, state of residence, time of day/night, weather conditions and the length of the trip.

(b) Some people might expect the probability of being injured in a short off-highway trip, or one in light traffic, to be less, and not want to bother with the seat belt in such a case. Statistics may prove such trips are more dangerous. (I don't know)

- Probability problems -
**drwls**, Wednesday, May 21, 2008 at 6:32am1. Multiply the probabilities of the separate outcopmes for each.

The probability that both will fail is (0.02)^2 = 0.0004

The probability that #1 will fail and #2 will not is 0.02*0.98 = 0.196

The probability that #2 will fail and #1 will not is also 0.01960

You need to add the last two results to get the probability that one will fail: 0.0392

The probability that neither will fail is (0.98)^2 = 0.9604

Note that the total is 0.0004 + 0.0392 + 0.9604 = 1.0000

2. The probability of surviving each trip is 1 - 4*10^-6 = 0.999996.

Take the 50,000 power of that as the probability of surviving 50,000 trips in a row with no fatality to anyone. That is 0.819. (The number seems too low - about 5% of people in the United States die in accidents; not 18%. But that is what you get using their numbers and "independent event" assumptions)

Trips are not really independent since the probability of a fatality on any trip varies with the driver's age, state of residence, time of day/night, weather conditions and the length of the trip.

(b) Some people might expect the probability of being injured in a short off-highway trip, or one in light traffic, to be less, and not want to bother with the seat belt in such a case. Statistics may prove such trips are more dangerous. (I don't know)

- Probability problems - explanation -
**drwls**, Wednesday, May 21, 2008 at 6:34amPlease excuse the multiple identical answers above. There was something wrong with Jiskha's operation this morning.

- Probability problems - explanation -
**Gayla**, Wednesday, May 21, 2008 at 9:06amThank you so very much for helping me with the two problems. I can now finish the rest of my homework. Take care and thanks for helping me and other students like me.

- Probability problems - explanation -

- Probability problems - explanation -
- Probability problems -
**khalid ahmed**, Saturday, July 19, 2008 at 3:49ami was probability mathematics trips to continuing of

- Probability problems -
**kutzi**, Friday, May 7, 2010 at 12:31amif you take 4 final exams randomly scheduled out of 10 possible times,how many ways can you be scheduled?

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