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805+ (Hard)|   Probability|      
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srikanth9502
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A string of lights is strung with red, blue, and yellow bulbs in a rat 13 Jul 2017, 20:54
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A string of lights is strung with red, blue, and yellow bulbs in a ratio of two to five to three, respectively. If 6 bulbs are lit in a random sequence, what is the probability that no color is lit more than twice?

A) 9/1000

B) 81/1000

C) 10/81

D) 1/3

E) 80/81
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B
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A string of lights is strung with red, blue, and yellow bulbs in a rat 14 Jul 2017, 11:59
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srikanth9502
A string of lights is strung with red, blue, and yellow bulbs in a ratio of two to five to three, respectively. If 6 bulbs are lit in a random sequence, what is the probability that no color is lit more than twice?

A) 9/1000

B) 81/1000

C) 10/81

D) 1/3

E) 80/81
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Dear srikanth9502,

I'm happy to respond. :-)

This is a brilliant question, a very clever variant of the standard "at least" scenario.

The only way no color will be lit more than twice is if we have exactly two of each of the three colors in our set of six. For instance, if we had less than two blues, we would have to have more than two reds or more than two whites. The only way we can have no color with more than two lights is if all three colors have exactly two lights. Thus, the question is really: what is the probability that a set of six has exactly two of each color?

Think about this one ordering: RRBBYY. The probability of this would be

probability = \(\tfrac{2}{10} \times \tfrac{2}{10} \times \tfrac{5}{10} \times \tfrac{5}{10} \times \tfrac{3}{10} \times \tfrac{3}{10}\) = \(\tfrac{10*10*3*3}{1000000}\) = \(\tfrac{9}{10000}\)

That's the probability of one ordering. All possible orderings would have the same probability, so we just need to multiply this by the number of orderings.

How many distinct ways can we put the letters RRBBYY in order? It's not 6!, because we have to divide out 2 for each pair of repetitions. (For example, we could put the two red bulbs in two different orders, and the arrangement would look the same.)

number of orders = \(\tfrac{6*5*4*3*2*1}{2*2*2}\) = 6*5*3 = 2*3*5*3 = 3*3*10 = 9*10

Now, multiply the probability we have by this number

final probability = \(\tfrac{9*9*10}{10000}\) = \(\tfrac{81}{1000}\)

Answer = (B)

Here are some more hard probability questions:
Challenging GMAT Math Practice Questions

Does all this make sense?
Mike :-)
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A string of lights is strung with red, blue, and yellow bulbs in a rat 14 Jul 2017, 20:00
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Thank you for the explanation.

I thought will be 2C2 x 5C2 X 3C2/10C6 (selection) multiplied by 6!/2!2!2!.

Can you please explain the error in my approach.
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A string of lights is strung with red, blue, and yellow bulbs in a rat 15 Jul 2017, 03:19
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mike

probability = \(\tfrac{2}{10} \times \tfrac{2}{10} \times \tfrac{5}{10} \times \tfrac{5}{10} \times \tfrac{3}{10} \times \tfrac{3}{10}\) = \(\tfrac{10*10*3*3}{1000000}\) = \(\tfrac{9}{10000}\)
Wouldn't this be a case similar to choosing a marbel and putting it back in the box?

As the bulb once lit is no more a part of outcome I calcuated the probablity as below:

probability = \(\tfrac{2}{10} \times \tfrac{1}{9} \times \tfrac{5}{8} \times \tfrac{4}{7} \times \tfrac{3}{6} \times \tfrac{2}{5}\)

Please correct me.

mikemcgarry
srikanth9502
A string of lights is strung with red, blue, and yellow bulbs in a ratio of two to five to three, respectively. If 6 bulbs are lit in a random sequence, what is the probability that no color is lit more than twice?

A) 9/1000

B) 81/1000

C) 10/81

D) 1/3

E) 80/81
Dear srikanth9502,

I'm happy to respond. :-)

This is a brilliant question, a very clever variant of the standard "at least" scenario.

The only way no color will be lit more than twice is if we have exactly two of each of the three colors in our set of six. For instance, if we had less than two blues, we would have to have more than two reds or more than two whites. The only way we can have no color with more than two lights is if all three colors have exactly two lights. Thus, the question is really: what is the probability that a set of six has exactly two of each color?

Think about this one ordering: RRBBYY. The probability of this would be

probability = \(\tfrac{2}{10} \times \tfrac{2}{10} \times \tfrac{5}{10} \times \tfrac{5}{10} \times \tfrac{3}{10} \times \tfrac{3}{10}\) = \(\tfrac{10*10*3*3}{1000000}\) = \(\tfrac{9}{10000}\)

That's the probability of one ordering. All possible orderings would have the same probability, so we just need to multiply this by the number of orderings.

How many distinct ways can we put the letters RRBBYY in order? It's not 6!, because we have to divide out 2 for each pair of repetitions. (For example, we could put the two red bulbs in two different orders, and the arrangement would look the same.)

number of orders = \(\tfrac{6*5*4*3*2*1}{2*2*2}\) = 6*5*3 = 2*3*5*3 = 3*3*10 = 9*10

Now, multiply the probability we have by this number

final probability = \(\tfrac{9*9*10}{10000}\) = \(\tfrac{81}{1000}\)

Answer = (B)

Here are some more hard probability questions:
Challenging GMAT Math Practice Questions

Does all this make sense?
Mike :-)
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A string of lights is strung with red, blue, and yellow bulbs in a rat 16 Jul 2017, 19:29
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srikanth9502
Thank you for the explanation.

I thought will be 2C2 x 5C2 X 3C2/10C6 (selection) multiplied by 6!/2!2!2!.

Can you please explain the error in my approach.
Show more
origen87
mike

probability = \(\tfrac{2}{10} \times \tfrac{2}{10} \times \tfrac{5}{10} \times \tfrac{5}{10} \times \tfrac{3}{10} \times \tfrac{3}{10}\) = \(\tfrac{10*10*3*3}{1000000}\) = \(\tfrac{9}{10000}\)
Wouldn't this be a case similar to choosing a marbel and putting it back in the box?

As the bulb once lit is no more a part of outcome I calcuated the probablity as below:

probability = \(\tfrac{2}{10} \times \tfrac{1}{9} \times \tfrac{5}{8} \times \tfrac{4}{7} \times \tfrac{3}{6} \times \tfrac{2}{5}\)

Please correct me.
Show more
Dear srikanth9502 & origen87,

I'm happy to respond. :-)

Look at the text of the problem.
A string of lights is strung with red, blue, and yellow bulbs in a ratio of two to five to three, respectively.

How many total bulbs are there? Just because there's a 2:5:3 ratio does not mean that there are 10 and only 10 bulbs in total. That is the kind of concrete literalist thinking about ratios that the GMAT regularly punishes.

In fact, the total number of bulbs was not mentioned. All we are told is that there is a "string of bulbs," a phrase that implies a large number of lights, and we have no idea how long this string is--maybe 12 ft, maybe 100 ft, maybe a mile. The implication is that we are dealing with a large population of bulbs, a number of bulbs so large that, for all intents and purposes, picking doesn't change the ratios at all. If I pick one bulb, the probability of picking blue is 5/10 = 1/2, and even if I put that blue aside and pick from the remaining population, the probability is still essentially 1/2 of picking one more blue bulb. With populations, we no longer have to consider the distinction of picking with or without replacement: that's only meaningful in a group of relatively small size.

This is a somewhat more extreme example, but let's say if I pick a random human being, there is a 1/2 probability that I pick a male. Well, even if 20 male get together someplace, and as it were, eliminate themselves from selection pool and start looking at the probability of picking another male from the rest of humanity, there would still be a 1/2 probability of picking a male. The numbers of humans are so large that adding or subtracting another 20 doesn't make much of a different. Just for comparison, on average about 250 human people are born each minute and about 105 people die each minute on the global scale. In that context, 20 people is considerably less than the minute-by-minute error margin.

Of course, the population of bulbs is probably not as large as the total human population of 7+ billion, but the fact that no grand total was specified implies that it is very large--large enough that adding or subtracting a couple bulbs will not change the ratios in any meaningful way.

Does all this make sense?
Mike :-)
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A string of lights is strung with red, blue, and yellow bulbs in a rat 16 Jul 2017, 22:12
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mike

Thanks a lot, definitely helps :)

mikemcgarry
srikanth9502
Thank you for the explanation.

I thought will be 2C2 x 5C2 X 3C2/10C6 (selection) multiplied by 6!/2!2!2!.

Can you please explain the error in my approach.
origen87
mike

probability = \(\tfrac{2}{10} \times \tfrac{2}{10} \times \tfrac{5}{10} \times \tfrac{5}{10} \times \tfrac{3}{10} \times \tfrac{3}{10}\) = \(\tfrac{10*10*3*3}{1000000}\) = \(\tfrac{9}{10000}\)
Wouldn't this be a case similar to choosing a marbel and putting it back in the box?

As the bulb once lit is no more a part of outcome I calcuated the probablity as below:

probability = \(\tfrac{2}{10} \times \tfrac{1}{9} \times \tfrac{5}{8} \times \tfrac{4}{7} \times \tfrac{3}{6} \times \tfrac{2}{5}\)

Please correct me.
Dear srikanth9502 & origen87,

I'm happy to respond. :-)

Look at the text of the problem.
A string of lights is strung with red, blue, and yellow bulbs in a ratio of two to five to three, respectively.

How many total bulbs are there? Just because there's a 2:5:3 ratio does not mean that there are 10 and only 10 bulbs in total. That is the kind of concrete literalist thinking about ratios that the GMAT regularly punishes.

In fact, the total number of bulbs was not mentioned. All we are told is that there is a "string of bulbs," a phrase that implies a large number of lights, and we have no idea how long this string is--maybe 12 ft, maybe 100 ft, maybe a mile. The implication is that we are dealing with a large population of bulbs, a number of bulbs so large that, for all intents and purposes, picking doesn't change the ratios at all. If I pick one bulb, the probability of picking blue is 5/10 = 1/2, and even if I put that blue aside and pick from the remaining population, the probability is still essentially 1/2 of picking one more blue bulb. With populations, we no longer have to consider the distinction of picking with or without replacement: that's only meaningful in a group of relatively small size.

This is a somewhat more extreme example, but let's say if I pick a random human being, there is a 1/2 probability that I pick a male. Well, even if 20 male get together someplace, and as it were, eliminate themselves from selection pool and start looking at the probability of picking another male from the rest of humanity, there would still be a 1/2 probability of picking a male. The numbers of humans are so large that adding or subtracting another 20 doesn't make much of a different. Just for comparison, on average about 250 human people are born each minute and about 105 people die each minute on the global scale. In that context, 20 people is considerably less than the minute-by-minute error margin.

Of course, the population of bulbs is probably not as large as the total human population of 7+ billion, but the fact that no grand total was specified implies that it is very large--large enough that adding or subtracting a couple bulbs will not change the ratios in any meaningful way.

Does all this make sense?
Mike :-)
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A string of lights is strung with red, blue, and yellow bulbs in a rat 16 Jul 2017, 22:52
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mike - I have the same question as origen87 .

However, I am unable to comprehend your answer to the question.

If possible could you please explain this in mathematical equations as why the solution :\(\frac{2c2 * 5c2 * 3c2}{10c6}\)* 90 ways isn't working here.

The ratios are 2 : 5 : 3 even we say that actual numbers of bulb are 4 , 10, 6

the probability of choosing a red bulb does not vary if we increase the factor by 2, 3 or 4..

P(R) =\(\frac{2}{10}\)= \(\frac{4}{20}\) = \(\frac{6}{30}\) = \(\frac{1}{5}\)
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A string of lights is strung with red, blue, and yellow bulbs in a rat 16 Jul 2017, 23:21
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mikemcgarry
srikanth9502
Thank you for the explanation.

I thought will be 2C2 x 5C2 X 3C2/10C6 (selection) multiplied by 6!/2!2!2!.

Can you please explain the error in my approach.
origen87
mike

probability = \(\tfrac{2}{10} \times \tfrac{2}{10} \times \tfrac{5}{10} \times \tfrac{5}{10} \times \tfrac{3}{10} \times \tfrac{3}{10}\) = \(\tfrac{10*10*3*3}{1000000}\) = \(\tfrac{9}{10000}\)
Wouldn't this be a case similar to choosing a marbel and putting it back in the box?

As the bulb once lit is no more a part of outcome I calcuated the probablity as below:

probability = \(\tfrac{2}{10} \times \tfrac{1}{9} \times \tfrac{5}{8} \times \tfrac{4}{7} \times \tfrac{3}{6} \times \tfrac{2}{5}\)

Please correct me.
Dear srikanth9502 & origen87,

I'm happy to respond. :-)

Look at the text of the problem.
A string of lights is strung with red, blue, and yellow bulbs in a ratio of two to five to three, respectively.

How many total bulbs are there? Just because there's a 2:5:3 ratio does not mean that there are 10 and only 10 bulbs in total. That is the kind of concrete literalist thinking about ratios that the GMAT regularly punishes.

In fact, the total number of bulbs was not mentioned. All we are told is that there is a "string of bulbs," a phrase that implies a large number of lights, and we have no idea how long this string is--maybe 12 ft, maybe 100 ft, maybe a mile. The implication is that we are dealing with a large population of bulbs, a number of bulbs so large that, for all intents and purposes, picking doesn't change the ratios at all. If I pick one bulb, the probability of picking blue is 5/10 = 1/2, and even if I put that blue aside and pick from the remaining population, the probability is still essentially 1/2 of picking one more blue bulb. With populations, we no longer have to consider the distinction of picking with or without replacement: that's only meaningful in a group of relatively small size.

This is a somewhat more extreme example, but let's say if I pick a random human being, there is a 1/2 probability that I pick a male. Well, even if 20 male get together someplace, and as it were, eliminate themselves from selection pool and start looking at the probability of picking another male from the rest of humanity, there would still be a 1/2 probability of picking a male. The numbers of humans are so large that adding or subtracting another 20 doesn't make much of a different. Just for comparison, on average about 250 human people are born each minute and about 105 people die each minute on the global scale. In that context, 20 people is considerably less than the minute-by-minute error margin.

Of course, the population of bulbs is probably not as large as the total human population of 7+ billion, but the fact that no grand total was specified implies that it is very large--large enough that adding or subtracting a couple bulbs will not change the ratios in any meaningful way.

Does all this make sense?
Mike :-)
Show more
Thanks a lot mike.

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A string of lights is strung with red, blue, and yellow bulbs in a rat 16 Jul 2017, 23:22
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Leo8
mike - I have the same question as origen87 .

However, I am unable to comprehend your answer to the question.

If possible could you please explain this in mathematical equations as why the solution :\(\frac{2c2 * 5c2 * 3c2}{10c6}\)* 90 ways isn't working here.

The ratios are 2 : 5 : 3 even we say that actual numbers of bulb are 4 , 10, 6

the probability of choosing a red bulb does not vary if we increase the factor by 2, 3 or 4..

P(R) =\(\frac{2}{10}\)= \(\frac{4}{20}\) = \(\frac{6}{30}\) = \(\frac{1}{5}\)
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Pls refer to the second explanation by mike

Sent from my Lenovo A7010a48 using GMAT Club Forum mobile app
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A string of lights is strung with red, blue, and yellow bulbs in a rat 17 Jul 2017, 09:42
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Leo8
mike - I have the same question as origen87.

However, I am unable to comprehend your answer to the question.

If possible could you please explain this in mathematical equations as why the solution :\(\frac{2c2 * 5c2 * 3c2}{10c6}\)* 90 ways isn't working here.

The ratios are 2 : 5 : 3 even we say that actual numbers of bulb are 4 , 10, 6

the probability of choosing a red bulb does not vary if we increase the factor by 2, 3 or 4..

P(R) =\(\frac{2}{10}\)= \(\frac{4}{20}\) = \(\frac{6}{30}\) = \(\frac{1}{5}\)
Show more
Dear Leo8,

I'm happy to respond. :-)

Yes, as we increase the numbers, the probability of picking a single bulb of any color stays the same. Of course, that's not the issue here. As soon as we are picking more than one bulb without replacement, things get very different. Think about it this way. Suppose there were just 10 bulbs. What is the probability, when picking without replacement, that the first three picked are all red? Zero! It's impossible because with 10 bulbs, we would have just two red bulbs, so it never would happen that we could get more than two. If we increase the number of bulbs to 20 or 40 or whatever, then it would be possible that the first three choices would be all red. We don't have to do a calculation: it's enough to see that it would be P = 0 with ten bulbs and P > 0 with a multiple of 10. Changing the number of bulbs changes the probability

Look at this expression.
\(\tfrac{2c2 * 5c2 * 3c2}{10c6}\)
This expression is explicitly assume that there are just 10 bulbs. Again, if there are just ten bulbs, then a different probability, the probability that all six chosen are red is zero, because, again, we only have two reds. But, if the number of bulbs is 100 or 1000 or something, then there would be a small probability that all six are red. Again, P = 0 with ten bulbs and P > 0 with 1000 bulbs. Of course, if the probability of one arrangement changes, that means the probabilities of all the other arrangements change, so the probability we are seeking would also change if we change from 10 to 1000 bulbs.

Finally, that expression is not even correct for the n = 10 case. It's hard to do take a 100% formula-based to probability. Probability is much more intuition-based. Calculating the n = 10 case would be a considerably harder problem: the solution would involve thinking through a large number of cases. It would be 10x harder than anything the GMAT would ask. The solution to that would be at least three-pages long and mind-boggling difficult: it can NOT be solved once and for all with a single formula.

Does all this make sense?
Mike :-)
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A string of lights is strung with red, blue, and yellow bulbs in a rat 28 Feb 2020, 08:23
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srikanth9502
A string of lights is strung with red, blue, and yellow bulbs in a ratio of two to five to three, respectively. If 6 bulbs are lit in a random sequence, what is the probability that no color is lit more than twice?

A) 9/1000

B) 81/1000

C) 10/81

D) 1/3

E) 80/81
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We see that the probability of getting a red bulb is 2/(2 + 5 + 3) = 1/5, that of getting a blue bulb is 5/10 = 1/2, and that of getting a yellow bulb is 3/10. Furthermore, if 6 bulbs are lit so that no color is lit more than twice, then there must be 2 red, 2 blue and 2 yellow bulbs. Therefore, the probability of getting 2 red, 2 blue and 2 yellow bulbs (in that order) is:

P(RRBBYY) = 1/5 x 1/5 x 1/2 x 1/2 x 3/10 x 3/10 = 9/10,000.

However, RRBBYY can be arranged in 6!/(2! x 2! x 2!) = 720/(2 x 2 x 2) = 90 ways.

Therefore, the probability of getting 2 red, 2 blue, and 2 yellow bulbs (in any order) is:

9/10,000 x 90 = 81/1000

Answer: B
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A string of lights is strung with red, blue, and yellow bulbs in a rat 21 Apr 2024, 18:25
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