# Independent odds, am I (+ friend) seeing this wrong or is there a mistake in the practice exam?

I found this exercise in a practice exam:

Any student has a 90% chance of entering a University. Two students are applying. Assuming each student’s results are independent, what is the probability that at least one of them will be successful in entering the National University?

A. $0.50$

B. $0.65$

C. $0.88$

D. $0.90$

E. $0.96$

I think the answer is something different than the answers above, namely $0.99$.

$0.01 = (0.1 \times 0.1)$ is the chance of neither, so $1 - 0.01$ must be $0.99$ right? But it's not part of the possible answers.

Other way: $(0.9 \times 0.9) + (0.9 \times 0.1) + (0.1 \times 0.9) = 0.99$

Am I missing something here?

• What if the university that gives them a 90% probability is not the National University? :P – immibis Aug 31 '17 at 5:54
• Please use MathJax in the future – gen-z ready to perish Aug 31 '17 at 11:31

Sometimes you have to challenge the options. Your concepts and answer are absolutely correct!

As the other answerers have noted, $0.99 = 1 - 0.1^2$ is indeed the correct answer.

As to what went wrong with the exercise, I suspect the problem statement has a typo, and the correct admission rate should be 80% instead of 90%. That would make option E ($0.96$ $=$ $1 - 0.2^2$) the correct one.

Why do I suspect that? It's because, as you note, for two independent trials with the same success rate, \begin{aligned} {\rm Pr}[\text{one succeeds}] &= 1 - {\rm Pr}[\text{both fail}] \\ &= 1 - (\text{failure rate})^2 \\ &= 1 - (1 - \text{success rate})^2, \end{aligned} which implies that, conversely, $$\text{success rate} = 1 - \sqrt{1 - {\rm Pr}[\text{one succeeds}]}.$$

Applying this "reverse" formula to the given options, we can work out what the admission rate would have to be for each option to be correct:

\begin{aligned} {\rm A}:\ {\rm Pr}[\text{one succeeds}] &= 0.5 &\implies& \text{success rate} = 1 - \sqrt{0.5}\phantom0 \approx 0.292893 \\ {\rm B}:\ {\rm Pr}[\text{one succeeds}] &= 0.65 &\implies& \text{success rate} = 1 - \sqrt{0.35} \approx 0.408392 \\ {\rm C}:\ {\rm Pr}[\text{one succeeds}] &= 0.88 &\implies& \text{success rate} = 1 - \sqrt{0.12} \approx 0.653589 \\ {\rm D}:\ {\rm Pr}[\text{one succeeds}] &= 0.9 &\implies& \text{success rate} = 1 - \sqrt{0.1}\phantom0 \approx 0.683772 \\ {\rm E}:\ {\rm Pr}[\text{one succeeds}] &= 0.96 &\implies& \text{success rate} = 1 - \sqrt{0.04} = 0.8 \end{aligned}

Out of these options, E is the only one where the probability of both students failing to be admitted works out to a nice square ($0.04 = 0.2^2$). For any of the options A to D to be (exactly) correct, the admission rate would have to be a very awkward irrational number.

Of course, it's also possible that the error is in the options themselves, and that the author of the exercise meant to include 0.99 as one of the options. The actual explanation might even be a combination of both possibilities — perhaps whoever wrote the exercise started with an admission rate of 80%, came up with a suitable answer set, and then later decided to change the admission rate to 90% but forgot to update the answers.

• good effort here – J Tg Aug 31 '17 at 10:38
• Good effort. Just wanted to add that you missed investigating one possibility: 90% was correct, but they intended to ask something else - exactly one student or both students getting admitted. But we can rule out that since none of the probabilities match, and thus your answer is correct. – Hans Olsson Aug 31 '17 at 14:35
• Note also that 0.88 is a tempting wrong answer, given 80% – Brondahl Aug 31 '17 at 15:22
• @Brondahl, I'm usually good at seeing where students went wrong or how they might go wrong to get a "likely" wrong answer, but I can't see what would make $0.88$ likely in this case.... – Wildcard Aug 31 '17 at 21:38
• Extremely crude .. if the student doesn't really understand probability at all and just sees 2 8s. – Brondahl Aug 31 '17 at 23:09

That's the standard way of doing it:

$P(\text{At Least One}) = 1 - P(\text{Both Fail}) = 1 - 0.1 \times 0.1 = 1 - 0.01 = 0.99$

And the not so standard way:

$P(1\text{ in}) + P(2\text{ in Without }1) = 0.9 + (0.9 \times 0.1) = .99$.

Or

\begin{align} P(1\text{ in Without }2) + P(2\text{in Without }1) + P(1\text{ in And }2\text{ in}) &= 0.9 \times 0.1 + 0.9 \times 0.1 + 0.9 \times 0.9 \\ &= 0.09 + 0.09 + 0.81 = .99 \end{align}

No matter how we cut it, you are right. They are wrong.

(It's probably just a typo.)

An important concept when setting multiple choice exams is to provide likely wrong answers. That is the examiner ideally wants to know if the candidate has mastered/remembered the concept. If typical mistakes are not one of the given options then what will be tested is the candidate's ability to re-evaluate their working in the face of evidence that it is wrong. Although this is also worthwhile testing I would suggest that in this case the examiner was just trying to test the concept of combining independent probabilities.

Hence, among the possible answers provided, there should be values like

Pr(success) * Pr(success);
Pr(success) * Pr(fail)
Pr(fail) * Pr(fail)
Pr(success) + Pr(success)             [despite this possibly being > 1]
Pr(fail) + Pr(fail)                   [despite this possibly being > 1]
Pr(success)


AND

1 - any of those above                [despite some possibly being < 0]


Of all the 19 "nice" probabilities 0.5, 0.1, 0.15 , 0.2 ,... 0.95 only 0.8 includes the correct answer as Ilmari Karonen pointed out. However the listed answers don't include any of the possible wrong answers for Pr(Success) = 0.8.

If instead of exact matches to the answers we include a difference of +/- 0.01 (that is the last decimal place shown) then the exact or approximate answer being in the shown options A, B, C, D, E for Pr(success) includes the cases 0.3, 0.4, 0.7, 0.8. Of those Pr(success) = {0.3, 0.7} have 2 approximate wrong answers in the list.
For Pr(success) = 0.7 we have
 Pr(success)*Pr(success) ~ 0.5 (answer A); and 1 - Pr(success)*Pr(success) ~ 0.5 (answer A) 

For Pr(success) = 0.3 we have
 Pr(fail)*Pr(fail) ~ 0.5 (answer A); and 1 - Pr(success)*Pr(success) ~ 0.9 (answer D) 

= = =

Any-way we look at it the question is a poor example of good testing and even with a single correction

• Pr(success) one of {0.3, 0.7, 0.8}; or
• answer E 0.96 => 0.99
it is doubtful that this test item would show good discrimination between good candidates and poorer candidates.

Ian

• God I love the internet sometimes – Expat C Nov 23 '17 at 6:08