hard probability math ?

Question

A student is trying to pass a competency exam. Each time she takes the exam she has a 15% chance of passing, and she is allowed a maximum of three attempts.
Draw a tree diagram to represent her attempts to pass the exam.


How many outcomes does your tree show?
What is the probability she will eventually pass the exam?
What is the probability she will take the exam three times?
What is the expected number of times she will take the exam?
What is the expected number of times she will fail the exam?

Answer 1

.................... / ....... \
................ 0.85 ..... 0.15 (Pass test 1)
.............. / ....... \
.......... 0.85 ..... 0.15 (Pass test 2)
........ / ....... \
... 0.85 ..... 0.15 (Pass test 3)
(Fail all 3)

1) How many outcomes does your tree show?

There are four outcomes:
P(Pass Test 1) = 0.15 = 15%
P(Pass Test 2) = 0.85 * 0.15 = 12.75%
P(Pass Test 3) = 0.85 * 0.85 * 0.15 = 10.8375%
P(Fail all 3) = 0.85 * 0.85 * 0.85 = 61.4125%

2) What is the probability she will eventually pass an exam.

This is the sum of all outcomes where she passes:
P(Pass Test 1) + P(Pass Test 2) + P(Pass Test 3)
= 38.5875%

Alternatively you could take the 1 - P(Fails all 3) and get the same answer.

3) What is the probability she will take the exam three times?

It didn't say whether she passes or fails, so the chance she has to take the test a third time is 0.85 * 0.85 = 72.25%

4) What is the expected number of times she will take the exam?
15% of the time, she'll take it once
12.75% of the time, she'll take it twice
72.25% of the time, she'll take it three times.

So the expected number of times is:
0.15 * 1 + 0.1275 * 2 + 0.7225 * 3 = 2.5725

So she's expected to take it 2.5725 times.

5) What is the expected number of times she will fail the exam?
15% of the time she will pass test 1 (failures = 0)
12.75% of the time she will pass test 2 (failures = 1)
10.8375% of the time she will pass test 3 (failures = 2)
61.4125% of the time she will fail all 3 (failures = 3)

0.15 x 0 + 0.1275 x 1 + 0.108375 x 2 + 0.614125 x 3 = 2.186625

So she's expected to fail it 2.186625 times.

Answer 2

4 outcomes with following probs...

Pass on try 1 = 15%
Pass on try 2 = 85% * 15% = 12.8%
Pass on try 3 = 85% * 85% * 15% = 10.8%
Don't pass at all = 85% * 85% * 85% = 61.4%

Prob she will pass eventually = 15%+12.8%+10.8% = 38.6%
Prob she will take the test three times = 1 - (15% + 12.8%) = 72.2%
Expected # of times = 15% * (1) + 12.8% * (2) + 72.2% * (3) = 2.6
Expected # of failures = 15% * (0) + 12.8% * (1) + 10.8% * (2) + 61.4% (*3) = 2.2