Question 1092403
<br>If the guesses are random, he has a 50% chance of getting the right answer on each problem.  We can make a "probability vector" for each question
{{{(1/2)Y+(1/2)N)}}}
to represent the fact that on each question his chances are 1 out of 2 to get the right answer ("Y") and 1 out of 2 to get the wrong answer ("N").<br>
Then we can multiply this expression by itself as many times as there are questions to determine the probabilities of getting certain numbers of answers right.  For a simple example, if there are just 2 questions, then we have
{{{((1/2)Y+(1/2)N)^2 = (1/4)Y^2 +(1/2)YN + (1/4)N^2}}}
which indicates a probability of 1/4 of getting both answers right or both wrong, and a probability of 1/2 of getting one right and one wrong.<br>
For a test with 10 questions, we have
{{{((1/2)Y+(1/2)N)^10}}}<br>
Note that this is equivalent to
{{{((Y+N)^10)/2^10 = ((Y+N)^10)/1024}}}
We need to find the probability that he gets at least 7 of the 10 answers right.  We can do that by expanding the expression using the binomial theorem.  But, since the binomial has both coefficients equal to 1, we can just use Pascal's Triangle.<br>
The 10th row of Pascal's Triangle, consisting of the coefficients in the expansion of (Y+N)^10, is<br>
1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1<br>
These numbers tell us that
P(he gets 10 right) = 1/1024
P(he gets 9 right) = 10/1024
P(he gets 8 right) = 45/1024
P(he gets 7 right) = 120/1024<br>
So the probability that he gets at least 7 answers right is 176/1024 = 11/64.