P & C problem

[abcgmat]

1. A certain junior class has 1000 students and a certain senior class has 800 students. Among these students there
are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each
class , what is the probability that 2 students selected will be sibling pair
1) 3/40,000 2)1/3,600 3)9/2,000 4)1/60 5)1/15
[spoiler]OA:A[/spoiler]
60/1000 * 1 /800 = 3/40,000
here we are not considering the situation of selecting one of the 60 from 800 students and 1 out of 1000 students: 60/800 * 1/1000

2. A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously
from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.
[spoiler]OA: D[/spoiler]
For option B we are considering two scenarios
D*(notD) + (non) D * D = n/10*(10-n)/9 + 10-n/10 * n/9
When do we need to consider only 1 scenario and when do we need to consider two scenarios

[Anurag@Gurome]

1. A certain junior class has 1000 students and a certain senior class has 800 students. Among these students there
are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each
class , what is the probability that 2 students selected will be sibling pair
1) 3/40,000 2)1/3,600 3)9/2,000 4)1/60 5)1/15
[spoiler]OA:A[/spoiler]
60/1000 * 1 /800 = 3/40,000
here we are not considering the situation of selecting one of the 60 from 800 students and 1 out of 1000 students: 60/800 * 1/1000

Required probability = No. of favorable outcome/ Total no. of outcomes
Total no. of outcomes = 800C1 * 1000C1
There are 60 sibling pairs, so no. of favorable outcomes = 60
Therefore, Required probability = = 60/(800*1000) = [spoiler]3/40,000[/spoiler]

The correct answer is A.

[Anurag@Gurome]

2. A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously
from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.
[spoiler]OA: D[/spoiler]
For option B we are considering two scenarios
D*(notD) + (non) D * D = n/10*(10-n)/9 + 10-n/10 * n/9
When do we need to consider only 1 scenario and when do we need to consider two scenarios

(1) The probability that the two bulbs to be drawn will be defective is 1/15.
Now the probability of drawing 2 defective bulbs out of a total of 10 bulbs depends on the number of defective bulbs, n. So, we can find value of n if we are given the probability; SUFFICIENT.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.
If there were 3 defective and 7 good bulbs or 7 defective and 3 good bulbs, then the probability of drawing one defective and one good bulb would be the same for both cases, so the given probability of 7/15 of drawing one defective and one good bulb will give 2 values of n, one less than 5 and another more than 5, but as we are given that n < 5, so we can get a unique value of n < 5; SUFFICIENT.

The correct answer is D.

[abcgmat]

I am still confused on When do we need to consider only 1 scenario and when do we need to consider two scenarios by looking at problem.

Why are we not considering two scenarios,
Selecting one student out of 1000 students and selecting 1(pair for the one selected) out of remaining 800 +
Selecting one student out of 800students and selecting 1(pair for the one selected) out of remaining 1000

You have used two different approaches for the two problems,and hence i am not able to link.
could you explain why the order doesnot matter for the above question and matters for defective bulb problem

[Bill@VeritasPrep]

The big difference is that #1 has two distinct groups (juniors and seniors) while #2 has one group (lightbulbs in the box).

By looking at #1 and starting with each group, you're effectively double-counting the probability. You can end up with the same pair whether you start with the junior or senior, so by counting both you're counting each sibling pair twice.

[1947]

The big difference is that #1 has two distinct groups (juniors and seniors) while #2 has one group (lightbulbs in the box).

By looking at #1 and starting with each group, you're effectively double-counting the probability. You can end up with the same pair whether you start with the junior or senior, so by counting both you're counting each sibling pair twice.

Bill, I agree with you that in #1 with each group we are double counting probability and that we will end up with same pair.
My question is how can we find in which scenario what to do ...in first go I had also approached thw question like the original poster.
Thanks in advance.