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Probability - Order / No Order

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Probability - Order / No Order

by vikram4689 » Tue Apr 17, 2012 8:06 am
I did 2 ques each in 2 ways and found that there is some discrepancy. Why both methods work for one question but not for another

1) from a standard deck (52 cards, 4 of each type), the probability that we get two Kings,

Method 1: Prob = 4C2/52C2 = 1/(13*17)
Method 2: 4 ways the first card is a King, 3 ways the second: 4*3 = 12 ways in total.
There are 52*51 ways to get two cards, in order. 12/(52*51) = 1/(13*17)is the probability

2) A bag has 6 red marbles and 4 marbles. What are the chances of pulling out a red and blue marble.
Method 1: # of ways of picking 1 red and 1 blue marble = (6c1)(4c1) = 6 x 4 = 24;
# of ways of picking 2 marbles in general = 10c2 = 45.
therefore, probability = 24/45
Method 2: 6 ways for red and 4 ways for blue = 24
total ways = 10*9
Prob = 24/90
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by Birottam Dutta » Tue Apr 17, 2012 8:18 am
In the second question, you also have to consider the equal probability of first taking out a blue marble followed by the red marble and the probability in this case is also 24/90.

Hence, total probability = 24/90+24/90 = 24/45.

Hope this clears your doubt.
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by vikram4689 » Tue Apr 17, 2012 8:23 am
ok, but why in the following problem we are not doing prob(1st senior then junior) + prob(1st junior then senior) for following problem

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 the 2 students selected will be a sibling pair ?

probability = 60/800*1/1000 = 3/40000
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by vikram4689 » Tue Apr 17, 2012 10:19 pm
Experts please clarify how above ques is different from earlier 2 ( which are clear to me)
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by aneesh.kg » Tue Apr 17, 2012 10:44 pm
Let's take the second question first.
Method 2: (6C1/10C1)*(4C1/9C1) is WRONG because this is P(Red)*P(Blue) which means that you are picking a Red Ball first and a Blue Ball next, whereas the question allows you to pick a (R&B) or (B&R). So, we have to arrange R and B in 2! factorial ways. The probability above will become 2! times to take care of this arrangement.
So, Answer: (6C1/10C1)*(4C1/9C1)*2! = 24/45
Since, it is possible for one to forget about sometimes, the Method 1 is better.

Method 1: (6C1)(4C1)/(10C2) = (6/10)(4/9)2! and this is correct.
Notice that the '2' of 10C2 goes on the numerator to take care of the arrangement. So, you don't need to worry about it anymore.

Let's come to the first question now.

The question is asking us to pick two kings. So (K1,K2) or (K2,K1) is the same thing.
Method 2: If you go in a certain order, it is OK because as I said (K1,K2) or (K2,K1) is the same thing. We don't need to multiply it by any factorial.
Method 1: This method is wonderful, because it takes of the fact that arrangement is not required without us worrying about anything.
Last edited by aneesh.kg on Tue Apr 17, 2012 10:49 pm, edited 1 time in total.
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by aneesh.kg » Tue Apr 17, 2012 10:47 pm
Lets' look at this from the definition of probability point of view.

Q2. We can pick a Red Ball first & a Blue Ball second OR a Blue Ball first & a Red Ball second.

P = P(R)*P(B) + P(B)*P(R)
P = 6/10*4/9 + 4/10*6/9 = [(6*4)/(10*9)]*2
Last edited by aneesh.kg on Tue Apr 17, 2012 10:57 pm, edited 1 time in total.
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by vikram4689 » Tue Apr 17, 2012 10:55 pm
Thanks Aneesh, how sibling question can be solved using arrangement method. shouldn't i do like R and B ball ques (i.e) multiply by 2
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by aneesh.kg » Tue Apr 17, 2012 11:05 pm
In the sibling question, the favourable outcomes are of selecting a pair which can be done in 60C1 ways. The question is not concerned with the arrangement inside that sibling pair, i.e. whether it is a S,J or a J,S.

So, there is no need to multiply with 2.
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by aneesh.kg » Tue Apr 17, 2012 11:12 pm
Remember the following fundamentals to solve:

- Probability = (No. of favourable events)/(Total no. of possible events)
- AND is multiplication
- OR is addtion

All the three questions given above can be solved by the above fundamentals.

1. P(2 Kings) = (No. of ways of selecting 2 kings)/(No. of ways of selecting any two cards)

2. P(one Red, one Blue) = (No. of ways of selecting a Red AND No. of ways of selecting a Blue)/(No. of ways of selecting two marbles)

3. P(one pair) = (No. of ways of selecting a pair)/(No. of ways of selecting a Senior AND No. of ways of selecting a Junior)
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by vikram4689 » Wed Apr 18, 2012 12:20 am
actually i am not able to differentiate b/w sibling and ball ques. IN both of them we need a PAIR

Sibling - senior and junior
Ball - read and blue

So why in one case we need to multiply by 2 and in other there is no such requirement
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by aneesh.kg » Wed Apr 18, 2012 1:04 am
No, The marbles/ball question doesn't require you to pick a pair of Red and Blue marbles. There is a subtle difference between saying that 'One of the two balls should be Red and other should be Blue' and saying that 'A pair of balls is required'. When you talk about a selecting pair, that pair becomes One single object and the arrangement inside that pair is irrelevant. The marbles are not paired up, but the siblings are. That's why the marbles have to arranged, whereas the siblings don't have to be.

Had the siblings question said that 'Select a Senior Sibling and his/her Junior Sibling', we'd have done the same thing as in marbles.
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by vikram4689 » Wed Apr 18, 2012 2:28 am
No, The marbles/ball question doesn't require you to pick a pair of Red and Blue marbles. When you talk about a selecting pair, that pair becomes One single object and the arrangement inside that pair is irrelevant.
But arrangement for current ball question is also irrevalent as you described in earlier posts, where am i getting confused
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by aneesh.kg » Wed Apr 18, 2012 2:38 am
No, arrangement for the ball question is not irrelevant. The balls have to be arranged but not the siblings.
Please read the explanation in my previous post again and carefully.
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by vikram4689 » Wed Apr 18, 2012 4:57 am
aneesh.kg wrote:No, arrangement for the ball question is not irrelevant. The balls have to be arranged but not the siblings.
Please read the explanation in my previous post again and carefully.
Above you are saying that arrangement is RELEVANT i.e. order matters BUT
(6C1/10C1)*(4C1/9C1) is WRONG because this is P(Red)*P(Blue) which means that you are picking a Red Ball first and a Blue Ball next, whereas the question allows you to pick a (R&B) or (B&R) So, we have to arrange R and B in 2! factorial ways.
This means arrangement or order is irrelevant coz above method,which is INCORRECT, gives prob of one of the cases (R then B)
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by GMATGuruNY » Wed Apr 18, 2012 5:04 am
vikram4689 wrote:actually i am not able to differentiate b/w sibling and ball ques. IN both of them we need a PAIR

Sibling - senior and junior
Ball - red and blue

So why in one case we need to multiply by 2 and in other there is no such requirement
When each selection is made from a DIFFERENT POOL -- one from a class of juniors, the other from a class of seniors -- then we DON'T have to worry about the ORDER of the selections.
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