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tough probability quest

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tough probability quest

by engg.manik » Sun Oct 04, 2009 9:10 am
Some toys include large, middle, and small model with red, yellow, green, or blue color. If numbers of all model-color combinations are the same, for example, number of red large toys is equal to number of green little toys. A boy wants a red-large toy. If his mother select one for him at random, what is the probability that at least one of the color and model will satisfy the boy?
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by cbenk121 » Sun Oct 04, 2009 8:59 pm
Probability = # of desired possibilities / total number of possibilities.

We're given three sizes, and four colors. So there's 12 possibilities.

The boy wants 1 possibility, a red large. So the probability is 1/12.

However, engg.manik possibly meant a red toy OR a large toy. In which case, there's three possibilities which involve a red toy, and four possibilities which involve a large toy, INCLUDING one already counted. So there's six possibilities that will satisify him, leading to a probability of 1/2.

List out to double check...

RL, RM, RS, LY, LG, LB, LR (ALREADY IN SET, SO DISCARD) = 6 possibilities.
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by Pooja Bhula » Wed Nov 25, 2009 10:33 pm
cbenk121 wrote:Probability = # of desired possibilities / total number of possibilities.

We're given three sizes, and four colors. So there's 12 possibilities.

The boy wants 1 possibility, a red large. So the probability is 1/12.

However, engg.manik possibly meant a red toy OR a large toy. In which case, there's three possibilities which involve a red toy, and four possibilities which involve a large toy, INCLUDING one already counted. So there's six possibilities that will satisify him, leading to a probability of 1/2.

List out to double check...

RL, RM, RS, LY, LG, LB, LR (ALREADY IN SET, SO DISCARD) = 6 possibilities.

I think The ans should be 1/12...The logic is that we want a red and large toy...therefore, the probability that it is a red toy = 1/3, the probability that it is a large toy = 1/4, so the probabilty that it is both red and large = 1/3 * 1/4 = 1/12... though i would want someone to confirm wat i am saying...
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by Stuart@KaplanGMAT » Wed Nov 25, 2009 11:55 pm
engg.manik wrote:Some toys include large, middle, and small model with red, yellow, green, or blue color. If numbers of all model-color combinations are the same, for example, number of red large toys is equal to number of green little toys. A boy wants a red-large toy. If his mother select one for him at random, what is the probability that at least one of the color and model will satisfy the boy?
cbenk has correctly interpreted the question:
what is the probability that at least one of the color and model will satisfy the boy?
So, any red toy is desirable, as is any large toy.

There are 4 possible large toys (1 of each colour) and 3 possible red toys (1 of each size). Here's where we have to be careful: we've counted the large red toy in both groups, so as cbenk noted, the total # of desired outcomes is 7-1=6, out of 12 total possible toys, to give us an answer of:

6/12 = 1/2

* * *

As an aside, please always post the answer choices for your questions, so we can better discuss strategic approaches to questions.
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by Pooja Bhula » Fri Nov 27, 2009 12:32 am
Thanks. I have understood the reasoning and the method. I just want to understand one more thing, while talking bout model - no. combinations the comparison of number of red large toys is made with number of green little toys. Does it mean that the toys have the same sizes in different colours? and does the '-' indicate 'or', for eg. in this case red or large? please help..
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by Stuart@KaplanGMAT » Fri Nov 27, 2009 1:59 am
Pooja Bhula wrote:Thanks. I have understood the reasoning and the method. I just want to understand one more thing, while talking bout model - no. combinations the comparison of number of red large toys is made with number of green little toys. Does it mean that the toys have the same sizes in different colours? and does the '-' indicate 'or', for eg. in this case red or large? please help..
"-" indicates "and" (in "If numbers of all model-color combinations are the same"); every toy has both a model and a colour. So a "red-large" toy is both red and large.
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by Pooja Bhula » Mon Nov 30, 2009 8:38 pm
If '-' means 'and' , then how come you have said that 'any red toy is desirable, as is any large toy' ? This means 'or' red or large..because you have not multiplied the no.s you have added them, so how is it 'and'?
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by Stuart@KaplanGMAT » Tue Dec 01, 2009 1:03 am
Pooja Bhula wrote:If '-' means 'and' , then how come you have said that 'any red toy is desirable, as is any large toy' ? This means 'or' red or large..because you have not multiplied the no.s you have added them, so how is it 'and'?
Because the question reads:
what is the probability that at least one of the color and model will satisfy the boy?
If we were to do this purely by brute force, we could just list all the toys:

red small
red middle
red large
yellow small
yellow middle
yellow large
blue small
blue middle
blue large
green small
green middle
green large

So we see there are 12 possibilities.

Now if we list the ones that have AT LEAST ONE of the qualities "red" or "large", we have:

red small
red middle
red large
yellow large
blue large
green large

So there are 6 desired outcomes.

So the final answer is 6/12 = 1/2

However, we don't need to brute force the question if we note that there are 3 reds and 4 larges. The point I was making about subtracting 1 from that total is because one of the toys is both red AND large, so we've counted it twice.
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by Pooja Bhula » Tue Dec 01, 2009 6:29 am
Stuart Kovinsky wrote:
Pooja Bhula wrote:If '-' means 'and' , then how come you have said that 'any red toy is desirable, as is any large toy' ? This means 'or' red or large..because you have not multiplied the no.s you have added them, so how is it 'and'?
Because the question reads:
what is the probability that at least one of the color and model will satisfy the boy?
If we were to do this purely by brute force, we could just list all the toys:

red small
red middle
red large
yellow small
yellow middle
yellow large
blue small
blue middle
blue large
green small
green middle
green large

So we see there are 12 possibilities.

Now if we list the ones that have AT LEAST ONE of the qualities "red" or "large", we have:

red small
red middle
red large
yellow large
blue large
green large

So there are 6 desired outcomes.

So the final answer is 6/12 = 1/2

However, we don't need to brute force the question if we note that there are 3 reds and 4 larges. The point I was making about subtracting 1 from that total is because one of the toys is both red AND large, so we've counted it twice.

Ok..I think i get it... thank you...
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