A certain freshman class has 600 students and a certain sophomore

For a given freshman, there will be only one partner sophomore and not 75. say for example:

a pairs with b
c pairs with d

If you've picked a from the freshmen group, b is the only partner from the sophomore group you can choose to pair with a. So, Probability of picking b out of a group of 900= 1/900.

Same for c and d or any other freshman-sophomore pair

75/600 * 1/900

1/7200

D

gmatpapa, kudos.

Even I had the same doubt some time back. Here's how I started differentiating between the two question types.

1. Probability of picking 1 red and 1 black out of the box containing 10 red and 10 black balls without replacement.
10/20*10/19+10/20*10/19

The orders mattered because we picked two balls from the box that contained mix of black and red.

In the question above, there are two disparate groups. You go to sophomore class and pick one AND then go to freshmen and pick another.

So:
1. The pupils from both year are not mixed here.
2. You clearly know that you are going to pick sophomore first and then the freshman.

OA is correct.

Hope the distinction is clear now!!!

Ahhh.. I think I get it now.. So when we have 2 dependent events (as in the example you mentioned) we will have to include both possibilities. And not in the case where we have two independent events.

Is that right, fluke?

No, I am afraid!!!

The distinction is not entirely based of the dependent or independent event. It is rather based of the fact that in the ball-pick scenario, we have all the balls mixed up and in the students-pick scenario, the groups are not mixed; they are separate groups.

e.g.
Probability of picking 1 red and 1 black out of the box containing 10 red and 10 black balls WITH replacement(independent events)

"10/20*10/20+10/20*10/20"
See the changes here. The ratio of favorable choices to total increased here but we still need to consider the two orders for we are picking randomly out of a mixed group.

I am not an expert. My theories are developed from the questions I encountered over a period of time. If you think there is a flaw, please challenge.

I'm getting a hang of what you're saying fluke.. Let me try out some more similar problems.. Let me put my new learning to test

Here is how i approached it.

there are 600*900 two people sets. (total outcomes)

75 sets are 2 people sets and lab partners ( favorable outcomes)

so the probability that the 2
students selected will be a set of lab partners = \(75/(600*900)\)

= 1/7200

Answer is D.

gmatpapa: As per your request, let me try to explain you the logic here.

Look at the question: "what is the probability that the 2
students selected will be a set of lab partners"

So you see 2 students in front of you - one from freshmen class and the other from sophomore. What is the probability that they are lab partners? There are only 75 pairs of lab partners possible. Total possible pairs are 600*900 (600C1 * 900C1)
So probability = 75/54000 = 1/7200

When you have a question similar to 'There are 4 boys and 3 girls. In how many ways can you make a pair of one boy and one girl?', what do you do? You say we can pick a boy in 4 ways and a girl in 3 ways so total 12 ways.

_ _
B G

There is one spot for a boy and one for a girl. Their positions are defined. As if there were two chairs, a blue one and a pink one for the boy and the girl respectively. So you don't arrange them. You fill the blue chair and the pink one. You don't say "we pick the boy first and then the girl or girl first and then the boy so we get 2 cases, right?"
It is a situation similar to this situation.

Can someone explain why its not 75C1*75C1/(600C1*900C1)
Thanks

how many sets possible

= 600*900

how many sets are there as lab partners

= 75

answer

= 75/540000

Though I couldn't solve it myself but explanation above helped

Is there only 1 set of 75? 2 ?

Once you select one of the 75 people in the freshmen class who are a part of a lab partner duo, there is only one way in which you can select his particular lab partner from the sophomore class. You are not allowed to pick any one of the 75 people in the sophomore class who are lab partners. You have to pick his lab partner only. Hence it is

75C1*1/(600C1*900C1)

You have to imagine that if you choose a person from the group in freshman (probability of 75/600 or 1/8) you only can match with 1 person on the other class, because the pair is exclusive.

Therefore:
1/8*1/900 = 1/7200

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