Question

In a certain true or false exam, there are 10 questions

set for the candidate. If a candidate guesses the answer

at each time

a) What is the probability that the candidate will get 8 or

more correct answers?

b) What is the probability that the candidate will answer

6 questions correctly?

c) What is the mean number of correct answers you

would expect the candidate to obtain? Find the

variance

Answer 1

a. n= total number of ways =2¹⁰ =1024

Since each answer can be true or false.

and m= favorable number of ways $C^8_{10}+C^{9}_{10}+C^{10}_{10}=\frac{10!}{8!2!}+\frac{10!}{9!}{1!}+\frac{10!}{10!0!}=45+10+1=56$

P=m/n=56/1024=7/128

b. $m=C^6_{10}=\frac{10!}{6!4!}=210$

P=210/1024=0.21

c. $\mu=\sum P(x)x$

$P(0)=P(10)=\frac{10!}{0!10!}/1024=1/1024$

$P(1)=P(9)=\frac{10!}{1!9!}/1024=10/1024$

$P(2)=P(8)=\frac{10!}{2!8!}/1024=45/1024$

$P(3)=P(7)=\frac{10!}{3!7!}/1024=120/1024$

$P(4)=P(6)=\frac{10!}{4!6!}/1024=210/1024$

$P(5)=\frac{10!}{5!5!}/1024=252/1024$

$\mu=1/1024(0+10)+10/1024(1+9)+45/1024(2+8)+120/1024(3+7)+210/1024(4+6)+(5)252/1024=5$

$\sigma_x^2=\sum P(x)x^2-(\sum P(x)x)^2=\\=1/1024(0+100)+10/1024(1+81)+45/1024(4+64)+120/1024(9+49)+210/1024(16+36)+252(25)/1024-25=27.5-25=2.5$