A certain junior class has 1,000 students and a certain

A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings 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 the 2 students selected at will be a sibling pair?

A. 3/40,000
B. 1/3,600
C. 9/2,000
D. 1/60
E. 1/15

Hi chetan2u,

I'm getting a bit confused here.
So in this question we can pick 60 from any one of the 2 groups, so 60/(800) * 1/1000 so we get the correct answer.

But we don't divide this by 2 factorial which I understand because either we pick senior first junior next or vice versa and then add both and then divide it by 2 because its the same.

Another question, : There are 4 distinct pair of siblings, what are the possible number of cases to form a committee of 4 with no siblings:

Here I got the answer by using 1- (exactly one sibling + exactly 2 sibling ), but that was a bit lengthy, ans = 16

so another method that results in the same answer here is the foll: 8* 6 * 4 * 2/ 4!, here we divide by 4 factorial since the order doesn't matter and we get the correct answer 16.

I understand that we divide by 4! because we have essentially formed a permutation and since order doesn't matter the number of cases reduce,

but if that's the case, then why arent we dividing by 2! here ( in the above question)

Check out this post: https://www.gmatclub.com/forum/veritas- ... er-matter/
It discusses a very similar situation.

Also, whether order matter or not depends on how you perceive it. In probability, you calculate P(A) = P(Favorable Outcomes)/P(Total Outcomes)

Where applicable, you can make the order matter in both numerator and denominator or not make it matter in both numerator and denominator. The answer will be the same.

This Question:

Order Matters:
Favorable outcomes = 60*1 + 60*1
Total outcomes = 1000*800 + 800*1000
P(A) = 3/40,000

Order doesn't matter:
Favorable outcomes = 60*1
Total outcomes = 1000*800
P(A) = 3/40,000

Another case: A bag has 4 balls - red, green, blue, pink (Equal probability of selecting each ball)
What is the probability that you pick two balls and they are red and green?

Order Matters:
Favorable outcomes = 2 (RG, GR)
Total outcomes = 4*3
P(A) = 1/6

Order doesn't matter:
Favorable outcomes = 1 (a red and a green)
Total outcomes = 4C2 = 6
P(A) = 1/6

Previous case: 2 red and 3 white balls. What is the probability that you pick two balls such that one is red and other is white?
In how many ways can you pick a red and a white ball? Here the probability of picking each ball is different. So we cannot cannot use our previous method. Here the probability of picking 2 red balls is not the same as that of picking a red and a white. Hence you have to consider the order to find the complete probability of picking a red and a white.

1) under what conditions ordering is considered? Pls correct if i m 100% correct? Pls add the correct approach.
KarishmaB GMATNinja Bunuel Elite097 avigutman ExpertsGlobal
MartyTargetTestPrep

a) if selection is from different baskets /groups (independent events) -no ordering ,
if from same baskets (impacts probability of selection of other marble/dependent events)- ordering considered

b) if final outcome is universally same (here siblings can't be swapped) - no ordering
if final outcomes are different via different selection techniques (Red first then White, or White first then Red) - ordering considered

There are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.

What is the probability of choosing ANY sibling from junior class? \(\frac{60}{1000}\) (as there are 60 of them).

What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be \(\frac{1}{800}\) (as there is only one sibling pair of chosen one).

So the probability of that the 2 students selected will be a sibling pair is: \(\frac{60}{1000}*\frac{1}{800}=\frac{3}{40000}\)

Answer: A.

This problem can be solved in another way:

In how many ways we can choose 1 person from 1000: \(C^1_{1000}=1000\);
In how many ways we can choose 1 person from 800: \(C^1_{800}=800\);
So total # of ways of choosing 1 from 1000 and 1 from 800 is \(C^1_{1000}*C^1_{800}=1000*800\) --> this is total # of outcomes.

Let’s count favorable outcomes: 1 from 60 - \(C^1_{60}=60\);
The pair of the one chosen: \(C^1_1=1\)
So total # of favorable outcomes is \(C^1_{60}*C^1_1=60\)

\(Probability=\frac{Number \ of \ favorable \ outcomes}{Total \ number \ of \ outcomes}=\frac{60}{1000*800}=\frac{3}{40000}\).

Answer: A.

Let’s consider another example:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

The same way here:

What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).

So the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000

Another way:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.

Favorable outcomes:
1 from 120=120C1=120
The pair of the one chosen=1C2=2
So total favorable outcomes=120C1*1C2=240

Probability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000

Hi @Bunuel, 

Why dont you include the other case as well where a Senior is picked first followed by a junior --> 60/800 * 1/1000.

Why doesn't your answer include this case?­

Answer A

Let's say you're picking out of the Junior class first and the senior class second (although the order doesn't make any difference). There are 1000 juniors and 60 of them have a sibling in the senior class, so you have a shot of choosing one of the siblings. Then you move onto the senior class. There are 800 seniors and only one sibling of the person you chose from the junior class. Thus, you have a chance of choosing the sibling.

Multiply the two equations together and simplify...and there's your answer.