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1. 6. A software company sets exams for programmers who wish to qualify to use their packages. Past records show that 55% of candidates taking the exam for the first time will pass, 60% of those taking it for the second time will pass, but only 40% of those taking the exam for the
third time will pass. Candidates are not allowed to sit the exam more than three times.

A programmer decides to keep taking the exam until he passes or is allowed no further
attempts. Find the probability that he will
(a) pass the exam on his second attempt, got 0.27
(b) pass the exam. got 0.892
Another programmer already has the qualification.
(c) Find, correct to 3 significant figures, the probability that she passed first time. got 0.617

At a particular sitting of the exam there are 400 candidates.
The ratio of those sitting the exam for the first time to those sitting it for the second time to those sitting it for the third time is 5 : 3 : 2

(d) How many of the 400 candidates would be expected to pass? nearest person.

for this, i added the ratio and got 10, the 400/10 =40, so in fact the ratio of people doing the exam was 200: 120: 80

then i * each value of the ratio and got 200*0.55 = 110 people
then i done 120*(0.45*0.6) =32.4, but the mark scheme says, 120*0.6 =72, but why, surely it is 120*(0.45*0.6) as to be able to take the test a second time, they need to fail the second time, so why would you not take account of the first fail probability of 0.45?

Thanks
2. (Original post by Hi, How are you ?)

then i done 120*(0.45*0.6) =32.4, but the mark scheme says, 120*0.6 =72, but why, surely it is 120*(0.45*0.6) as to be able to take the test a second time, they need to fail the second time, so why would you not take account of the first fail probability of 0.45?

Thanks
Because those that are taking it for the second time have already failed the first time. The probability of them passing this time is 0.6, hence 120*0.6.
3. (Original post by ghostwalker)
Because those that are taking it for the second time have already failed the first time. The probability of them passing this time is 0.6, hence 120*0.6.
So why would you not include the 1st fail probability?
4. Because they have already failed. That event has already occurred so it is certain. -> If Person A is doing the test for the second time, what is the probability they failed the first time? = 100%.
5. (Original post by Hi, How are you ?)
So why would you not include the 1st fail probability?
Because for that 120 people who are taking it for the second time, there is no question of whether they will succeed or fail for the first time. They failed.

These 120 people are those that failed the first test (the ones that passed the first time aren't there - they don't need to resit it).

I feel I'm not saying anything different, so perhaps someone else with a different slant on things may be able to help.
6. (Original post by ghostwalker)
Because for that 120 people who are taking it for the second time, there is no question of whether they will succeed or fail for the first time. They failed.

These 120 people are those that failed the first test (the ones that passed the first time aren't there - they don't need to resit it).

I feel I'm not saying anything different, so perhaps someone else with a different slant on things may be able to help.
So because they have already failed the first time, the 0.45 probability is not included as we already know they failed the 1st time, so we are trying to find out weather they would pass of fail, right?

i kind of get it, but still can't see to get my head around the fact of not using the 0.45 :/
7. It's just that since they've already failed once, the probability that they DID fail is 1.
Just like if i missed school yesterday, the probability that i missed school is 1, and that's independent of whether i miss school tomorrow. Dunno if that random analogy helped at all :/
8. (Original post by JackS94)
It's just that since they've already failed once, the probability that they DID fail is 1.
Just like if i missed school yesterday, the probability that i missed school is 1, and that's independent of whether i miss school tomorrow. Dunno if that random analogy helped at all :/
So in fact, since they DID fail, you times by 1, (but you just miss that out) as the probability they fail is 100% in their first go, right?
9. (Original post by Hi, How are you ?)
So in fact, since they DID fail, you times by 1, (but you just miss that out) as the probability they fail is 100% in their first go, right?
That's how i'd understand it
10. Just considering these 120 who are taking the test for the second time.

We're told that the probability that any one of them will pass on this their second attempt is 0.6. Hence the 120*0.6.

If we do 120*0.45*0.6, we're treating these 120 as if they haven't sat the test before, and that they will fail on their first attempt (i.e. this one they're doing now), and then pass on the next resit.

But these 120 have already sat the first test and failed it. This is their second attempt.

Don't know if that helps, but looks like JackS94 has found something that works for you.
11. Thanks guys =)

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