Solution :
a). The probability that the student will [tex]\text{correctly answer}[/tex] the 1st question after the 4th attempt.
P (correct in the 4th attempt)
= [tex]$(1-0.75)^3 \times 0.75$[/tex]
= 0.01171875
b). The probability that the student will [tex]\text{correctly answer}[/tex] 3 questions after 10 total attempts.
P( X = 3) for X = B in (n = 10, p = 0.75)
= [tex]$C(10,30) \times 0.75^3 \times 0.25^7$[/tex]
= 0.0031
c). The mean and the standard deviation for the number of attempts up to when the students gets all the questions correct is :
There are = 6 success, p = 0.75.
Therefore, this is a case of a negative binomial distribution.
[tex]$E(X)=\frac{k}{p}$[/tex]
[tex]$=\frac{6}{0.75}$[/tex]
= 8
So, [tex]$\sigma = \frac{\sqrt{k(1-p)}}{p}$[/tex]
[tex]$\sigma = \frac{\sqrt{6(1-0.75)}}{0.75}$[/tex]
= 1.6330
