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March 30th, 2016, 09:23 AM   #1
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There are 10 houses on the street where you live. On a snowy day, the probability that someone will shovel the driveway by hand is 65%. What is the probability that less than 4 driveways will be shoveled by hand?

There are 8 houses on the street where you live. On a snowy day, the probability that someone will shovel the driveway by hand is 60%. What is the probability that more than 4 driveways will be shoveled by hand?

The probability that a person owns an iPAD is 0.32. The probability that, given the person owns an iPad, he/she attends Fairview is 0.12. What is the probability that a person owns an iPAD and attends Fairview?
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March 30th, 2016, 09:47 AM   #2
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Assuming that each shovelling is independent, for 1) and 2), you can apply the formula for the pmf of a binomial distribution. For the first one, the probability is $\displaystyle 0.35^{10} + \displaystyle{10 \choose 1}0.65^1 \times 0.35^9 + \displaystyle{10 \choose 2}0.65^2 \times 0.35^8 + \displaystyle{10 \choose 3}0.65^3 \times 0.35^7$. I think you can do the other one now.

For the third question, let I be the event that the person owns an iPad and F be the event that he attends Fairview. P(I) = 0.32, P(F|I) = 0.12. P(I ∩ F) = P(I)P(F|I) = 0.32 * 0.12 = 0.0384.
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March 30th, 2016, 11:46 AM   #3
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Originally Posted by alymrod View Post
There are 10 houses on the street where you live. On a snowy day, the probability that someone will shovel the driveway by hand is 65%. What is the probability that less than 4 driveways will be shoveled by hand?
The probability that any one driveway will not be shoveled by hand is 1- 0.65= 0.35. The probability that none of the 10 driveways will be shoveled is $\displaystyle (0.35)^{10}$. The probability that any specific one driveway will be shoveled is $\displaystyle (0.65)(0.35)^9$. Because there are 10 different possible driveways, the probability that exactly one driveway shoveled is $\displaystyle 10(0.65)(0.35)^9$. The probability that any specific two driveways are shoveled is [math](0.65)^2(0.35)^8[math]. Those two could be any of the $\displaystyle \frac{10!}{2!8!}= 45$ pairs of driveways so the probability that exactly two driveways are shoveled is $\displaystyle 45(0.65)^3(0.35)^7$. Finally, the probability that any specific three of the ten driveways being shoveled is $\displaystyle (0.65)^3(0.35)^7$ and there are $\displaystyle \frac{10!}{3!7!}= 240$ sets of three driveways so the probability exactly three driveways are shoveled is [math]240(0.65)^2(0.35)^7. The probability of "less than four driveways" being shoveled is the sum of those four values.

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There are 8 houses on the street where you live. On a snowy day, the probability that someone will shovel the driveway by hand is 60%. What is the probability that more than 4 driveways will be shoveled by hand?
Same idea. Use 0.60 as the probability of any one driveway being shoveled and 0.40 as the probability of any one driveway not being shoveled.

Quote:
The probability that a person owns an iPAD is 0.32. The probability that, given the person owns an iPad, he/she attends Fairview is 0.12. What is the probability that a person owns an iPAD and attends Fairview?
Imagine 1000 people. Since "the probability that a person owns an iPAD is 0.32", (0.32)(10000)= 3200 people own an iPAD. Since "the probability that, given the person owns an iPAD, he/she attends Fairview is 0.12", (0.12)(3200)= 384 people who own an iPAD and attend Fairview so the probability that any one person both owns an kPAD and attends Fairview is 384/10000.
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March 30th, 2016, 06:06 PM   #4
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I am confused with the probability of less than four driveways being shoveled.
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March 31st, 2016, 11:24 AM   #5
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The probability of fewer than 4 driveways being shoveled is 0.0260242804964843.
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