Solution :
a). The probability that the student will
the 1st question after the 4th attempt.
P (correct in the 4th attempt)
=
![$(1-0.75)^3 * 0.75$](https://img.qammunity.org/2022/formulas/mathematics/college/p3ebqmyx3gqn38v35tyjg76eh7p8pztmbp.png)
= 0.01171875
b). The probability that the student will
3 questions after 10 total attempts.
P( X = 3) for X = B in (n = 10, p = 0.75)
=
![$C(10,30) * 0.75^3 * 0.25^7$](https://img.qammunity.org/2022/formulas/mathematics/college/5vqrpmmnzlq7z48x97zzdfrvyswpl7ziuw.png)
= 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.
![$E(X)=(k)/(p)$](https://img.qammunity.org/2022/formulas/mathematics/college/lhel2eey5ibc2s3936sacah1oy2i7ssati.png)
![$=(6)/(0.75)$](https://img.qammunity.org/2022/formulas/mathematics/college/9z1o2j25r6dcmwo55694oq98j448uabj7v.png)
= 8
So,
![$\sigma = (√(k(1-p)))/(p)$](https://img.qammunity.org/2022/formulas/mathematics/college/gjnzdjg6x9dpnopnxtql2wjvwmo50ygy3j.png)
![$\sigma = (√(6(1-0.75)))/(0.75)$](https://img.qammunity.org/2022/formulas/mathematics/college/aeqkm4v2oxwdf1vh3j0mfv6uqovu1hgn95.png)
= 1.6330