Question

A student is given a test with $10$ question of true and false type. If the gets $4$ or more questions correct, he is declared pass. Assuming that he guesses the answer to each questions, compute the probability that he will passes in the test.

Answer 1

Probability of student getting correct answer is $\frac{1}{2}$ and getting answer wrong is $\frac{1}{2}$

Let $P=P\left(correct\right)=\frac{1}{2}$

$q=P\left(wrong\right)=\frac{1}{2}$

For passing the exam student need to get atleast is answers correct

By using Bermaulli's triangle

$x=$ no. of success trial

$x=4,5,6,7,8,9,10$

Probability of passing the exam $P\left(X\right)=P\left(4\right)+P\left(5\right)+P\left(6\right)+P\left(7\right)+P\left(8\right)+P\left(9\right)+P\left(10\right)$

${{}^{n}C}_x{\left(p\right)}^x{\left(q\right)}^{n-x}$

$P(X=x)={{}^{10}C}_x{\left(\frac{1}{2}\right)}^x{\left(\frac{1}{2}\right)}^{10-x}={{}^{10}C}_x{\left(\frac{1}{2}\right)}^{10}$

for $X=4,5,6,7,8,9,10$

$P(X=x)={\left(\frac{1}{2}\right)}^{10}[{{}^{10}C}_4+{{}^{10}C}_5+{{}^{10}C}_6+{{}^{10}C}_7+{{}^{10}C}_8+{{}^{10}C}_9+{{}^{10}C}_{10}]={\left(\frac{1}{2}\right)}^{10}[{{}^{10}C}_6+{{}^{10}C}_6+{{}^{10}C}_5+{{}^{10}C}_7+{{}^{10}C}_8+{{}^{10}C}_9+{{}^{10}C}_{10}]$

$={\left(\frac{1}{2}\right)}^{10}[2\times \frac{10!}{6!4!}+\frac{10!}{5!5!}+\frac{10!}{7!3!}+\frac{10!}{8!2!}+\frac{10!}{9!1!}+\frac{10!}{0!10!}]$

$={\left(\frac{1}{2}\right)}^{10}[2\times 210+252+120+45+10+1]$

Probability of passing exam $=\frac{848}{2^{10}}=\frac{848}{1024}=\frac{106}{128}=\frac{53}{64}$