Probability of 4 driveways being shoveled by hand

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itsjustme

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[FONT=verdana, geneva, lucida, lucida grande, arial, helvetica, sans-serif]I had a similar question with 10 houses, but I can't remember how to set it up...
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[FONT=verdana, geneva, lucida, lucida grande, arial, helvetica, sans-serif]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?[/FONT]
 
itsjustme said:
I had a similar question with 10 houses, but I can't remember how to set it up...



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?
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What are your thoughts?

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Assuming that each shovelling is independent, for 1) and 2), I applied the formula for the pmf of a binomial distribution. For the other one the probability is 0.3510+(101)0.651×0.359+(102)0.652×0.358+(103)0.653×0.3570.35101010.6510.3591020.6520.3581030.6530.357. now the question is different but only by 2 different numbers asking for 8 houses and 60% I'm just confused.


 
itsjustme said:
Assuming that each shovelling is independent, for 1) and 2), I applied the formula for the pmf of a binomial distribution. For the other one the probability is 0.3510+(101)0.651×0.359+(102)0.652×0.358+(103)0.653×0.3570.35101010.6510.3591020.6520.3581030.6530.357. now the question is different but only by 2 different numbers asking for 8 houses and 60% I'm just confused.
Click to expand...
Try \(\displaystyle \displaystyle\sum\limits_{k = 5}^8 {\binom{8}{k}{{\left( {.6} \right)}^k}{{\left( {.4} \right)}^{8 - k}}} \)
 
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