Slight harder probability question. Trying to help my son on a probability question. Got 0.2511 for part a, but not quite the answer. For part b, can't even understand the question :-p
To qualify as a driver, a candidate must first pass a theory test and then pass a practical test. If a candidate passes the theory test, they are allowed two attempts to pass the practical test. If they fail the practical test at both attempts, they must start again by taking a new theory test.
The following table shows the proportion of candidates that pass these tests at their first, second, third, fourth, fifth and sixth attempts.

*

*Theory: 65% (1st), 72% (2nd), 80% (3rd), 85% (4th), 91% (5th), 96%
(6th)


*Practical: 35% (1st), 50% (2nd), 65% (3rd), 80% (4th), 95% (5th), 99%
(6th)
Questions:
a) A particular candidate passes her first theory test. Find the exact probability that she has to take a theory test exactly once more before qualifying.
b) Find the exact probability that a candidate qualifies on the fourth test that he takes.

 A: Part A 
It is given that the candidate has passed first theory test. We need the following sequence:

*

*Fail the first practical test. $Pr = 0.65$

*Fail the second practical test. $Pr = 0.5$

*Pass the second theory test. $Pr = 0.72$

*Either pass the third practical test; or fail the third but pass the fourth. 
$Pr = 0.65+0.35*0.8 = 0.93$
Cumulative Probability = $0.65*0.5*0.72*0.93 = 0.21762 $
Part B
Possible sequences are:
Theory Fail, Theory Fail, Theory Pass, Practical Pass
Theory Fail, Theory Pass, Practical Fail, Practical Pass
Any other sequence will take more than or less than 4 tests.
Hence, calculate the probability of these sequences, add them up and get the answer.
A: a) In order to qualify after taking exactly one more theory test, note that the candidate has already taken and passed the first theory test, so must fail both of her practical test attempts which occurs with probability $P(P_1^c)P(P_2^c)$ where $P_i$ denotes passing the practical test on the $i^{th}$ attempt. Then, to qualify after the second theoretical attempt they have to pass that attempt, and either pass the 3$^{rd}$ or 4$^{th}$ practical. Thus, the probability desired is
$P(P_1^c)P(P_2^c)P(T_2)P(P_3) + P(P_1^c)P(P_2^c)P(T_2)P(P_3^c)P(P_4)$ = $P(P_1^c)P(P_2^c)P(T_2)[P(P_3) + P(P_3^c)P(P_4)]$ = $.65*.5*.72*[.65 + .35*.8]$
A: Got the answer for part a:
0.65 x 0.5 x 0.72 (0.65 + 0.35 x 0.8) = 0.21762
Explanation:

*

*Failed the 1st Practical test = 1 - 0.35 = 0.65

*Failed the 2nd Practical test = 1 - 0.5 = 0.5

*Passed the 2nd Theory test = 0.72

From this point onward, there are two possible routes:

*

*4a) Pass the 3rd Practical test = 0.65

*4b) Fail the 3rd Practical test and pass the 4th Practical test = (1 - 0.65) x 0.8

Add up the two probabilities to get the final answer!
Answer for part b by whoisit:
A bit elaboration of the question: what is the probability for a candidate to take exactly four tests (any combinations of Theory Test and Practical Test) to qualify as a driver.
Two possible combinations are:

*

*Fail the 1st Theory Test = 1 - 0.65 = 0.35

*Fail the 2nd Theory Test = 1 - 0.72 = 0.28

*Pass the 3rd Theory Test = 0.8

*Pass the 1st Practical Test = 0.35

or

*

*Fail the 1st Theory Test = 1 - 0.65 = 0.35

*Pass the 2nd Theory Test = 0.72

*Fail the 1st Practical Test = 1 - 0.35 = 0.65

*Pass the 2nd Practical Test = 0.5

Probability = 0.35 x 0.18 x 0.8 x 0.35 + 0.35 x 0.72 x 0.65 x 0.5 = 0.10934
Answers given at the back of the book:

Thank you everyone for helping in the question!
(Sorry my reputation score is still too low to cast an up vote)
