Lights

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Lights

by Ashujain » Fri Jun 22, 2012 2:08 am
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) 9/10000
B) 81/1000
C) 10/81
D) 1/3
E) 80/81

OA: B

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by Anurag@Gurome » Fri Jun 22, 2012 5:22 am
Ashujain wrote: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?
The ratio of number of red, blue and yellow bulb is 2:5:3
Hence, there are 2 red bulbs, 5 blue bulbs, and 3 yellow bulbs in (2 + 5 + 3) = 10 bulbs

Hence,
  • Probability of picking one red bulb = 2/10 = 1/5
    Probability of picking one blue bulb = 5/10 = 1/2
    Probability of picking one yellow bulb = 3/10
As no color is lit more than twice and there are a total of 6 bulbs of three colors, there must 2 bulbs of each color. Now 2 bulbs of each color can be arranged differently.

Number of arrangements of 6 bulbs in which each two are identical = 6!/[(2!)*(2!)*(2!)] = (6*5*4*3*2)/(2*2*2) = (6*5*3) = 90

Now for each of this 90 arrangements we have to pick a bulb of each color twice.
Hence, probability of getting any such arrangement = [(1/5)^2]*[(1/2)^2]*[(3/10)^2] = [(1/5)*(1/2)*(3/10)]^2 = (3/100)^2 = 9/10000

Hence, required probability = 90*(9/10000) = 81/1000

The correct answer is B.
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by GMATGuruNY » Fri Jun 22, 2012 8:46 am
Ashujain wrote: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) 9/10000
B) 81/1000
C) 10/81
D) 1/3
E) 80/81

OA: B
To satisfy the given ratio, let the 10 lights be RR-BBBBB-YYY.

There are 6 positions in the sequence and 3 colors.
No color can be lit more than twice.
The implication is that each color must be lit EXACTLY twice: RRBBYY.
Here's why:
If any of the 3 colors is lit FEWER THAN 2 times -- if there is only one R, for example -- then one of the remaining colors will have to be lit MORE THAN 2 times: RBBBYY or RBBYYY.

Question rephrased: What is the probability that each color is lit EXACTLY TWICE?

One way for each color to be lit exactly twice: RRBBYY
Since there are 2 red bulbs, the number of ways to get RR in the first two positions = 2*2.
Since there are 5 blue bulbs, the number of ways to get BB in the next two positions = 5*5.
Since there are 3 yellow bulbs, the number of ways to get YY in the last two positions = 3*3.
To combine these options, we multiply:
Number of ways to get RRBBYY = 2*2*3*3*5*5 = 900.

Total possible ways:
RRBBYY is only ONE way to get 2 of each color.
We must account for ALL of the ways to get 2 of each color.
Any arrangement of RRBBYY will include 2 of each color.
Thus, the result above (900) must be multiplied by the number of ways to arrange RRBBYY.
The number of ways to arrange 6 elements = 6!.
When an arrangement includes identical elements, we must divide by the number of ways to arrange each set of identical elements.
The reason is that when identical elements swap positions, the arrangement is unchanged.
The number of ways to arrange RR = 2!
The number of ways to arrange BB = 2!.
The number of ways to arrange YY = 2!.
Thus:
The number of ways to arrange RRBBYY = 6!/(2!2!2!) = 90.

Total number of ways for each color to be lit exactly twice:
Multiplying the results above, we get:
900*90 = 81,000.

Total ways to light the 10 bulbs:
There are 6 positions in the sequence.
Since each position could be occupied by any one of the 10 bulbs, we get:
10*10*10*10*10*10 = 1,000,000.

P(2 of each color) = 81,000/1,000,000 = 81/1000.

The correct answer is B.
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by thestartupguy » Sat Jun 23, 2012 5:08 am
Anurag@Gurome wrote:
Ashujain wrote: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?
The ratio of number of red, blue and yellow bulb is 2:5:3
Hence, there are 2 red bulbs, 5 blue bulbs, and 3 yellow bulbs in (2 + 5 + 3) = 10 bulbs

Hence,
  • Probability of picking one red bulb = 2/10 = 1/5
    Probability of picking one blue bulb = 5/10 = 1/2
    Probability of picking one yellow bulb = 3/10
As no color is lit more than twice and there are a total of 6 bulbs of three colors, there must 2 bulbs of each color. Now 2 bulbs of each color can be arranged differently.

Number of arrangements of 6 bulbs in which each two are identical = 6!/[(2!)*(2!)*(2!)] = (6*5*4*3*2)/(2*2*2) = (6*5*3) = 90

Now for each of this 90 arrangements we have to pick a bulb of each color twice.
Hence, probability of getting any such arrangement = [(1/5)^2]*[(1/2)^2]*[(3/10)^2] = [(1/5)*(1/2)*(3/10)]^2 = (3/100)^2 = 9/10000

Hence, required probability = 90*(9/10000) = 81/1000

The correct answer is B.
Anurag, I did not understand the highlighted part. Could you please explain

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by Anurag@Gurome » Sat Jun 23, 2012 5:17 am
thestartupguy wrote:As no color is lit more than twice and there are a total of 6 bulbs of three colors, there must 2 bulbs of each color. Now 2 bulbs of each color can be arranged differently

...I did not understand the highlighted part. Could you please explain
In the random sequence chosen there are a total of 6 bulbs.
We know that there are three possible colors : red, blue, or yellow.

Now the condition is "no color is lit more than twice" among the six bulbs. Hence, the maximum number of red, blue or yellow bulbs in the sequence of 6 bulbs is 2.

Now, if you consider one bulb of any color is there, then there must be three bulbs of some other color. Hence, bulbs of each color must occur exactly twice.

Now the arrangements of this 6 bulbs (2 of each three colors) can be different.
For example, RRBBYY or RBRBYY or RYBYBR etc

Hope that helps.
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by Mo2men » Sat Jul 15, 2017 10:35 am
GMATGuruNY wrote:
Ashujain wrote: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) 9/10000
B) 81/1000
C) 10/81
D) 1/3
E) 80/81

OA: B
To satisfy the given ratio, let the 10 lights be RR-BBBBB-YYY.

There are 6 positions in the sequence and 3 colors.
No color can be lit more than twice.
The implication is that each color must be lit EXACTLY twice: RRBBYY.
Here's why:
If any of the 3 colors is lit FEWER THAN 2 times -- if there is only one R, for example -- then one of the remaining colors will have to be lit MORE THAN 2 times: RBBBYY or RBBYYY.

Question rephrased: What is the probability that each color is lit EXACTLY TWICE?

One way for each color to be lit exactly twice: RRBBYY
Since there are 2 red bulbs, the number of ways to get RR in the first two positions = 2*2.
Since there are 5 blue bulbs, the number of ways to get BB in the next two positions = 5*5.
Since there are 3 yellow bulbs, the number of ways to get YY in the last two positions = 3*3.

To combine these options, we multiply:
Number of ways to get RRBBYY = 2*2*3*3*5*5 = 900.
Dear Mitch,
In the red parts, can you shed light why you 3*3 & 5*5??
Thanks

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Re: Lights

by Scott@TargetTestPrep » Fri Feb 21, 2020 7:36 am
Ashujain wrote:
Fri Jun 22, 2012 2:08 am
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) 9/10000
B) 81/1000
C) 10/81
D) 1/3
E) 80/81

OA: B
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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