SOLUTION: A quiz cosists of 10 multiple-choice questions, each with 6 possible answers. For someone who makes random guesses for all of the answers, find the probability of passing if the mi
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Question 511144: A quiz cosists of 10 multiple-choice questions, each with 6 possible answers. For someone who makes random guesses for all of the answers, find the probability of passing if the minimum passing grade is 50 %.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
This one is easy if you would be satisfied with a wild approximation. A snowball in hell would have a better chance. Here's why.
Here we are faced with getting at least 5 out of 10 correct. That means we need to find the probability of getting exactly 5, then exactly 6, then 7...and so on up to 10, and then add these probabilities.
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
Since we need the sum of several of these calculations:
\ =\ \sum_{k\,=\,5}^{10}\left(10\cr\,k\right\)\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{10\,-\,k})
You can save yourself the calculation of one of the terms by realizing that the probability of something happening is equal to 1 minus the probability of it not happening and then performing the sum:
\ =\ 1\ -\ P_{10}\left(\leq4,\frac{1}{6}\right)\ =\ \sum_{k\,=\,0}^{4}\left(10\cr\,k\right\)\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{10\,-\,k})
That is only 5 terms instead of 6.
You can also save yourself a good deal of work if you realize that taking 1/6 to the 8th, 9th, and 10th power is going to result in some mighty big denominators and therefore some mighty small terms. In fact, the value of these three terms will be lost in any reasonable round-off of the numeric approximation of the answer. Hence:
\ \approx\ \sum_{k\,=\,5}^{7}\left(10\cr\,k\right\)\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{10\,-\,k})
Is close enough (only off by 2 in the 5th decimal place), and you only have to calculate 3 terms.
And finally, if all of this nasty arithmetic leaves you cold, and no one will ever think ill of you for feeling that way, open a spreadsheet, Excel on the PC or Numbers on the Mac, then pick a cell and enter the following formula:
})
and get your answer directly upon hitting the Enter key.
John
My calculator said it, I believe it, that settles it
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