Briana guesses at all 10 true/false questions on her history test. Find each probability.

P(exactly 6 correct)

Question

Briana guesses at all 10 true/false questions on her history test. Find each probability.
 
P(exactly 6 correct)

anonymous · Answer

6/20

anonymous · Answer

There are 10 questions, and 2 options for each question (true and false), so there are 20 options in total. So the probability of her choosing 6 of the correct options is 6 out of 20, which can then be simplified to 3/10 :)

anonymous · Answer

That's not it- you forget that there are multiple permutations for 6 answers correct/4 incorrect

anonymous · Answer

We might as well simplify Brianna to a mindless 50/50 robot and the answers to correct/incorrect:|dw:1346179735212:dw|

anonymous · Answer

This branch continues until the 10th question. Each branch doubles the number of points, so at the 10th (final) stage there are 2^10=1024 points. Now to work out which ones have 4 incorrect and 6 correct answers down the line ('upstream')from there.

anonymous · Answer

We might as well do this by working out how many ways it is possible to get 4 incorrect answers (for nicer numbers).

anonymous · Answer

In mathematical terms, it's 4 C 10 (same as 6 C 10), but I can't work out a way to explain it. Wait a minute.

anonymous · Answer

I can only explain it algebraically now. Call a correct answer= c and incorrect=n
Each are actually=0.5.
(c+n)^10=1
But if we expand this, we get (unsimplified) 1024 terms. How many of them are with 6 correct answers? Look at the 10th row of Pascal's triangle, 6 along (or 4 along, P.T. is symmetrical). This is the number which will be in front of c^6n^4 after simplification. It turns out to be 210 ([6 C 10] on a calculator).

All the terms, after the expansion of (c+n)^10 still=1 (as (c+n)^10=1), so 210*c^6n^4/total probability will be the prob. you are looking for (with T.P.=1, as explained) 
c=0.5, and n=0.5, and compute the answer!

Note that the result here is the same as dividing 210 (the no. of possible ways to get 6c/4n) by 1024 (total possible endpoints)

anonymous · Answer

Look up Pascal's triangle if the 210 seemed a little artificial, but the answer, in short, is just (no. of ways to get 6c)/(no. of possible ways to finish the test).