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Basic Probability

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Basic Probability

by hai1 » Tue Mar 09, 2010 3:01 am
1. In a class of 30 students, there are 17 girls and 13 boys. Five are A students, and three of these students are girls. If a student is chosen at random, what is the probability of choosing a girl or an A student?

2. Three cards are chosen at random from a deck without replacement. What is the probability of getting a jack, a ten and a nine

3. In a shipment of 100 televisions, 6 are defective. If a person buys two televisions from that shipment, what is the probability that both are defective?
Last edited by hai1 on Tue Mar 09, 2010 3:10 am, edited 2 times in total.
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by sanju09 » Tue Mar 09, 2010 3:07 am
hai1 wrote:In a class of 30 students, there are 17 girls and 13 boys. Five are A students, and three of these students are girls. If a student is chosen at random, what is the probability of choosing a girl or an A student?
A girl or an A student are not mutually exclusive events here. There are 17 girls in total, out of which 3 are A students. Hence, 17 girls plus 2 boys from the A group would form the favorable lot for this probability question, and the required probability = [spoiler]19/30[/spoiler].
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by sanju09 » Tue Mar 09, 2010 3:33 am
hai1 wrote:1. In a class of 30 students, there are 17 girls and 13 boys. Five are A students, and three of these students are girls. If a student is chosen at random, what is the probability of choosing a girl or an A student?

2. Three cards are chosen at random from a deck without replacement. What is the probability of getting a jack, a ten and a nine

3. In a shipment of 100 televisions, 6 are defective. If a person buys two televisions from that shipment, what is the probability that both are defective?
Please post one question per thread and learn why it is advisable.

2. It's a pure selection of 3 cards out of 52 that can be accomplished in 52C3 = 22100 ways. A jack, a ten and a nine can be had in 4*4*4 = 64 ways, and the required probability = 64/22100 = [spoiler]16/5525[/spoiler].

3. If the two televisions are bought simultaneously and NOT one after the another, as the language hiddenly suggests, then there are 100C2 ways of selecting two televisions for purchase, and there are 6C2 ways that the selected two televisions are from the defective lot. Hence, the required probability = 6C2/100C2 = [spoiler]1/330[/spoiler].
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by yeahdisk » Tue Mar 09, 2010 5:18 am
If there are 4 Jacks, 4 tens, 4 nines in every pack of cards (Would this kind of prior knowledge be assumed on the GMAST?)

anyway, IMO answer = [spoiler](4/52) * (4/51) * (4/50)[/spoiler]
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by sanju09 » Tue Mar 09, 2010 5:38 am
yeahdisk wrote:
sanju09 wrote:It's a pure selection of 3 cards out of 52
Not quite - there are 4 Jacks, 4 tens, 4 nines in every pack of cards. (Would this kind of prior knowledge be assumed on the GMAST?)

anyway, IMO answer = [spoiler](4/52) * (4/51) * (4/50)[/spoiler]
Two points to ponder here...

1. Oh yes! I missed the term "without replacement" in the stem.

2. Although the GMAT has witnessed very a few questions based on the prior knowledge of commonly played cards, but even then, GMAT assumes you know the basics associated with cards, chess, etc too.
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by yeahdisk » Tue Mar 09, 2010 6:44 am
actually I revisit my original answer:

My original answer of 4/52 * 4/51 * 4/50

I ths probability of getting a 9 - 10 - jack in that specific order.

there are 3! combinations of orders of getting those cards, so my final answer is

[spoiler]6 * (4/52 * 4/51 * 4/50)[/spoiler]
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by nysnowboard » Tue Mar 09, 2010 10:59 pm
Thanks guys! This stuff is invaluable... I know the basic theory but the devil is in the details so to speak. Keep it coming =)

Btw, if the television question had been re-worded as "successive purchases", would this be simply

6/100 x 5/99? (Assuming independence(you don't have a shady salesperson and/or extremely bad luck))


For the card problem, I also forgot the 3! possible ways of getting the three cards >_<...

Thank you!
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