Question

A student guesses the answers to 6 question on a true-false quiz. Find the probability that the indicated number of guesses are correct....At least 3

Answer 1

There's some trick to do this that makes it a lot simpler, but I'm forgetting it, so we're going to do it the long way.

First, notice that the question is equivalent to asking for the probability of getting at most 2 incorrect. This is a smaller number, so let's work with that instead.

Answer 2

The probability of getting 0 incorrect is \[\binom{6}{0}\cdot \left({1 \over 2}\right)^6\]The probability of getting 1 incorrect is\[\binom{6}{1}\cdot \left({1 \over 2}\right)^6\]And the probability of getting 2 incorrect is\[\binom{6}{2}\cdot \left({1 \over 2}\right)^6\]Let me know if you're unsure of where I got those formulas.

Answer 3

Now you just need to add them together.\[\binom{6}{0}\cdot \left({1 \over 2}\right)^6+\binom{6}{1}\cdot \left({1 \over 2}\right)^6+\binom{6}{2}\cdot \left({1 \over 2}\right)^6\]\[=\left({1 \over 2}\right)^6\cdot\left( \binom{6}{0}+\binom{6}{1}+\binom{6}{2}\right)\]\[=\left({1 \over 64}\right)\cdot (1+6+15)\]\[=\left({1 \over 64}\right)\cdot22\]\[={22 \over 64}\]\[={11 \over 32}\]

Answer 4

im kinda unsure of were you got those formulas. Can you break it down a little more for me please. Not so good with math, Thanks ! (:

Answer 5

Thanks for asking!

Let's start with the first one. We're looking for the probability that we get 0 incorrect/6 correct. The probability of getting a single question correct is \[{1\over2}\]The probability of getting 2 in a row correct is \[{1\over2}\cdot{1\over2}=\left({1 \over 2}\right)^2\]Therefore, the probability of getting 6 in a row correct/0 incorrect is \(\displaystyle \left({1 \over 2}\right)^6\)But why do we multiply by \(\displaystyle\binom{6}{0} \text{?}\) Because we need to choose 0 problems out of 6 to answer incorrectly!

Answer 6

Now what about the second one? You may be wondering why there's still a \[\left({1\over2}\right)^6\]even though we're not getting 6 correct answers. It boils down to the fact that we have to choose 1 problem out of 6 to answer incorrectly. So we can choose a certain problem to get incorrect, and multiply by \(\displaystyle\binom{6}{1}\).

So let's suppose we get the last one incorrect. Then we have \(\displaystyle \left({1\over2}\right)^5\) probability of getting the first 5 correct, and \(\displaystyle {1\over2}\) chance of getting the last one incorrect. If we multiply them together, we still get \(\displaystyle \left({1\over2}\right)^6\)!

Now we just need to multiply by \(\displaystyle\binom{6}{1}\). The the probability would be \[\binom{6}{1}\cdot \left({1 \over 2}\right)^6\]

Answer 7

By a very similar reasoning to above, we get \[\binom{6}{2}\cdot \left({1 \over 2}\right)^6\]for the last formula. We just need to choose 2 problems to answer incorrectly, and then multiply by \(\displaystyle \binom{6}{2}\)

Answer 8

Did that help you understand?

Answer 9

Yes now i understand it much better . Thanks for all the help (:

Answer 10

You're welcome.