certain junior class has 1000 students

[Guest]

This from Gmat prep1

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

The OA is A. However i got C (9/2,000) . The steps i followed is: (60/1000*60/800). Whats wrong with this approach?

[so19]

Assume the junior siblings in the 60 sibling pairs are numbered as Js(junior sibling)1, Js2, Js3, .... , Js60. And there are 1000 students in a junior class.
Also assume that the senior siblings in the 60 sibling pairs are numbered as Ss(senior sibling)1, Ss2, Ss3, .... , Ss60. And there are 800 students in a senior class.

We get to pick one from each class, a junior class and a senior class, and have to find the probability that the two will be a pair, i.e. (Js1, Ss1), (Js2, Ss2), ... , (Js60, Ss60).

Solution:
1. There are 60 possible siblings in a junior class to choose from = 60/1000 (here we do not care which sibling we pick as long as we pick the exact match in a senior class)

(Two different interpretations are possible from here)
2a. There are 60 siblings in a senior class and we must match our pick out of 60 the one we picked from the junior class = (60/800)*(1/60); or
2b. We do not care how many siblings in a senior class because we just have to pick the match out of 800 seniors = 1/800

1 * 2a = (60/1000)*(60/800)*(1/60) = 3/40000
1 * 2b = (60/1000)*(1/800) = 3/40000

Hope this helps.

[Guest]

So19...

Thanx for the great explanation. It really helped!!!

[divya]

so frankly, I am little confused here:

why isnt it that it can be case1 or case2
so 3/40000 + 3/40000 = 6/40000

I see 2 scenario's from Sol9's post, so I view it as P(a) + P(b) in case of mutually exclusive cases. Can anyone please clarify.

From sol9's post:
(Two different interpretations are possible from here)
2a. There are 60 siblings in a senior class and we must match our pick out of 60 the one we picked from the junior class = (60/800)*(1/60); or
2b. We do not care how many siblings in a senior class because we just have to pick the match out of 800 seniors = 1/800

1 * 2a = (60/1000)*(60/800)*(1/60) = 3/40000
1 * 2b = (60/1000)*(1/800) = 3/40000

[RonPurewal]

http://www.manhattangmat.com/forums/off ... t1211.html

you could also use the basic probability formula here: (# of successful outcomes) / (total # of possible outcomes).

the "successful outcomes" are the sibling pairs, of which we have been informed that there are 60.

the "possible outcomes" are the different ways of selecting ANY pair from the overall pool. this can be done in 800 x 1000 different ways, because a junior and a senior are being selected independently. (analogy: if you have 3 shirts and 4 pairs of pants, then you have 12 outfits. this is basically the same deal.)

this gives 60 / (800 x 1000), so, (a). the fraction is of course the same, but note that it has been derived by a method that is substantially different.

[link]

this is a good question.

pick one from junior class, the chance you pick one belongs to sibling is 60/1000=3/50

then pick another from senior class, the chance you pick on belong to sibling is 60/800=3/40

if the questions ask that is the chance both time u will pick someone from siblings, that is it - (3/50)*(3/40)

however, the questions goes to further step, asked you what is the probability that the 2 students selected will be a sibling pair. So there is a 1/60 chance u can pick the right guy. so the final answer is (3/50)*(3/40)*(1/60)=3/40,000

that is all.

[RonPurewal]

this is a good question.

pick one from junior class, the chance you pick one belongs to sibling is 60/1000=3/50

then pick another from senior class, the chance you pick on belong to sibling is 60/800=3/40

if the questions ask that is the chance both time u will pick someone from siblings, that is it - (3/50)*(3/40)

however, the questions goes to further step, asked you what is the probability that the 2 students selected will be a sibling pair. So there is a 1/60 chance u can pick the right guy. so the final answer is (3/50)*(3/40)*(1/60)=3/40,000

that is all.

that works, too.

[SEPY]

Can anyone pls answer divya's doubt :
so frankly, I am little confused here:

why isnt it that it can be case1 or case2
so 3/40000 + 3/40000 = 6/40000

I see 2 scenario's from Sol9's post, so I view it as P(a) + P(b)

Can we have situation where we select first from Junior section and then from senior section + first from senior section and then from Junior section.

In that case we have 3/40000 + 3/40000 = 6/40000

[mschwrtz]

Order doesn't matter here.

If you pick the senior and then the junior you get:
60/800*1/1000

If you pick the junior and then the senior you get:
60/1000*1/800

If you pick them simultaneously you get the same result.

But you ask, "Can't you get each pair in two different ways, S-J or J-S?" No. If you want to look at it that way, you need to say that your odds of drawing a twin on the first pull are 120/1800, and the odds of drawing his pair on the second are...well...NOT 1/1799, because you have to draw someone from the other class.

So you'll need something like a probability tree. But if you do it correctly you'll get the same result as above, albeit by the tortured route:

(120/1800)(60/120)(1/800)+(120/1800)(60/120)(1/1000)

Have fun with the fractions.

[supratim7]

you could also use the basic probability formula here: (# of successful outcomes) / (total # of possible outcomes).

Ron's method is surely the cleanest for this particular problem.
But I have a query; here it goes...

Case-1) A standard deck of cards contains 26 red and 26 black cards. If exactly 2 cards are selected from a standard deck (without replacement), what is the probability that 1 red and 1 black card are selected?

The probability that 1 red and 1 black card are selected = P(1st card RED) x P(2nd card BLACK) + P(1st card BLACK) x P(2nd card RED) = 26/52 x 26/51 + 26/52 x 26/51 = 26/51

In this case, the ORDER of selection MATTERS

Case-2) 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?

The probability that 2 students selected will be sibling pair = P(1st student JUNIOR) x P(2nd student SENIOR) = (60/1000) x (1/800) = 3/40000

In this case, the ORDER of selection DOESN'T MATTER

Why the ORDER of selection matters in Case-1 but not in Case-2? What am I missing/not considering??

Many thanks | Supratim

[jlucero]

Quite simply, because there's no overlap in the student problem. You are selecting one person from the junior class AND one from the senior class. These are two separate groups and you need to select exactly one from each group. There is no way to select a junior from the senior class and vice versa. Because you have to choose one from each group, it doesn't matter where you begin.

In the card problem, you are selecting cards from a deck, you could pick a red card OR a black card. So it does matter where you begin- if you choose a black card first, then you need to select a red card. While if you choose a red card first, then you need to select a black card. Which is why we have to count both of these possibilities.

[supratim7]

Aah.. I get it :) Thank you so much Joe.

[supratim7]

Would it be OK to generalize in following manner?

Because the *pool* is SAME in Case-1 but DIFFERENT in Case-2, the ORDER of selection matters in Case-1 but not in Case-2.

Many thanks | Supratim

[RonPurewal]

Would it be OK to generalize in following manner?

Because the *pool* is SAME in Case-1 but DIFFERENT in Case-2, the ORDER of selection matters in Case-1 but not in Case-2.

Many thanks | Supratim

that's pretty much how it works.

[supratim7]

Thank you Ron. Appreciate it :)