A certain junior class has 1000 students and a certain senio

[rschunti]

A certain junior class has 1000 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 will be a sibling pair?

A, 3/40000, B 1/3600, C. 9/2000, D, 1/60, E, 1/15

This is GMATPREP question. What is the best approach to solve this question?

[StaceyKoprince]

First, make sure you understand what you have to do. First, pick a student from one of the two classes who has a sibling in the other class. Then, pick that student's sibling in the other class.

Junior first: 60 students with siblings out of 1000 juniors = 60/1000.
That junior's senior sibling: 1 student out of 800 seniors = 1/800.

I want both conditions (junior sibling AND senior sibling), so I multiply: [60/1000] * [1/800] = 6/(100*800) = 6/80,000 = 3/40,000.

You could also start with the senior first: 60 students with siblings out of 800 seniors = 60/800
That senior's junior sibling: 1 student out of 1000 juniors = 1/1000
Multiply and get the same result as above: 3/40,000.

[gmatwork]

I have a doubt -

Why won't we add the probability of two cases - (Case I - Junior first then Senior) + (Case II Senior First then Junior)

= 60/1000 * 1/800 + 60/800 * 1/1000

What is wrong with this ?


In a similar question we did consider the number of ways - (example)


For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads 'at least' 4 times?

The answer to this question is
5[((0.6)^4)(0.4)] + (0.6)^5

P(atleast 4 heads) = P(4 heads) + P( 5 heads)

Now while solving this question when we calculate P(atleast 4 heads)-

P(4heads) = (0.6)^4)(0.4) * 4 - Here we multiply by four because there are four ways to get four heads and one tail

Thinking on the similar lines, I utilized the same logic for the question in discussion and considered both all the possible scenarios in which we can pick the required group.

I am not clear with respect to when to multiply with the number of ways a group can be formed or number of ways we can get four heads etc etc VERSUS when not to multiply by the number of ways---- in order to get the final answer in probability and combinatorics questions.

[tim]

your calculation of 60/1000 * 1/800 + 60/800 * 1/1000 takes into account the case of junior-then-senior as well as senior-then-junior. this approach will work, but you have to deal with the fact that you've overcounted because now each pair you've identified shows up twice in the final analysis. take your calculation and divide by 2, and you'll be fine..

[gmatwork]

Sorry, I still don't understand this point.

[RonPurewal]

Sorry, I still don't understand this point.

you only do what you're doing if you are making both selections from the same group. in this problem, we are selecting from two different pools, so that's not the correct approach.

for instance: (a normal american deck of cards contains 52 cards, of which 13 are hearts and 13 are spades.)

* if you draw two cards from the SAME DECK, the chance of picking one heart and one spade is
(Spade then Heart) + (Heart then Spade)
= (13/52)(13/51) + (13/52)(13/51)

* if you draw one card from DECK 1 and another card from DECK 2, then the probability that you get a heart from DECK 1 and a spade from DECK 2 is just (13/52)(13/52).
tim's point is that, in cases such as this one, you have to just pick one order of selection and stick with it. i.e., you just pick DECK 1 then DECK 2; you don't also pick in the reverse order.

--

you can also test your logic using cases for which the answers are already obvious with common sense.
for instance, consider 2 decks that contain 10 cards, ALL of which are black.
... what's the probability of picking a black card from each deck?
(obviously, the answer to this question is 1, since you have nothing but black cards.)

using your logic, you would get that the answer to this question is (1)(1) + (1)(1) = 2, which is not a valid probability.
instead, the answer is just (1)(1).

[gmatwork]

Thanks a lot, Ron. On the same line of thought one more question,

A basket has 5 apples - 1 bad and 4 good. If we were to pick two apples simultaneously, then what is the probability that one apple will be bad?

Is it ok to use conditional probability approach for this question? (considering that picking of apples is happening without replacement so when we are picking things simultaneous this should be the same as picking things without replacement ......is this understanding correct?)

When using conditional probability - this pool will be considered a single pool - right???

because both bad and good apples are part of the same basket and NOT two different baskets so I will consider -two separate cases (spoiled first then good) + (good first then bad)

I understand that there are easier ways to solve this one.

[tim]

there are lots of ways to solve this one, and conditional probability can work. can you take us through your calculations so i can let you know if you're doing it correctly?

[gmatwork]

Hello Tim and Ron,

With conditional probability it should be
P( good, bad) +P(bad, good)

1/5 *4/4 + 4/5*1/4 = 2/5 ( which is the correct ans)

My question again is that we are considering both cases since it is a single basket (although there are two subcategories of apples but since the probability of one event depends on the probability of the other and picking the apples form the same basket as opposed to two different baskets, we will have to consider both cases - considering either one of the two happens first in each case.

As opposed to the above in the following question we see -

If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end after the fourth throw?

Calculate the probability that either of the two sequences, THHH or HTTT, will occur. The probability of THHH occurring = (1/2)^4

The probability of THHH occurring = (1/2)^4 and then add the two

In this case when we calculate the probability of THHH we do not add up the probability of THHH , HTHH, HHTH, HHHT because essentially the probability one event does not impact the probability of the consecutive event so we do not need to consider the probability of possible permutations.

Although I can't see this from the perspective of elements belonging to the same group OR different groups - the approach that Ron used to explain me the concept of when I need to consider different cases and add them Vs. when not to add different cases (please refer to the above post by Ron in which he answered my question using examples)

[tim]

"With conditional probability it should be
P( good, bad) +P(bad, good)
1/5 *4/4 + 4/5*1/4 = 2/5 ( which is the correct ans)"

This is correct because you are making two DIFFERENT selections, namely the FIRST apple and the SECOND apple. When you make that distinction, you need to account for the possibility that you will pick good AND THEN bad as well as the possibility of bad AND THEN good. The AND THEN tells you to multiply the probabilities..

"If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end after the fourth throw?
In this case when we calculate the probability of THHH we do not add up the probability of THHH , HTHH, HHTH, HHHT because essentially the probability one event does not impact the probability of the consecutive event so we do not need to consider the probability of possible permutations."

No, the reason you don't need to add in HTHH, HHTH, and HHHT is because they don't work! They aren't situations that satisfy the conditions of the problem..

I see you've asked a lot of questions on the forums about combinatorics and probability. My best advice to you is to make sure you have a story in mind of how you are picking objects, stick to that, and use terms that accurately reflect what is happening. That said, keep the questions coming and we'll be glad to help!