Problem Solving

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 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

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.

Can somebody please help me point out where I am going wrong.
The soln to this question can be 2 red + 2 blue + 2 yellow arrangement.
RRBBYY can be arranged in 6!/2!*2!*2! = 90 ways.
2 red lights can be chosen from 2 red lights in 1 way (2C2).
2 blue lights can be chosen from 5 yellow lights in 5C2 ways.
And 2 yellow lights can be chosen from 3 in 3C2 ways.
Total ways 6 lights can be chosen from 10 lights is 10C6 ways.
Answer = (90*2C2*5C2*3C2)/10C6

I know my answer for this probability question is wrong since it doesnt match any of the answer choices but can someone please guide me.

Anju@Gurome wrote:

varun289 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

Can somebody please help me point out where I am going wrong.
The soln to this question can be 2 red + 2 blue + 2 yellow arrangement.
RRBBYY can be arranged in 6!/2!*2!*2! = 90 ways.
2 red lights can be chosen from 2 red lights in 1 way (2C2).
2 blue lights can be chosen from 5 yellow lights in 5C2 ways.
And 2 yellow lights can be chosen from 3 in 3C2 ways.
Total ways 6 lights can be chosen from 10 lights is 10C6 ways.
Answer = (90*2C2*5C2*3C2)/10C6

I know my answer for this probability question is wrong since it doesnt match any of the answer choices but can someone please guide me.

GMATGuruNY 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

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.

Can somebody please help me point out where I am going wrong.
The soln to this question can be 2 red + 2 blue + 2 yellow arrangement.
RRBBYY can be arranged in 6!/2!*2!*2! = 90 ways.
2 red lights can be chosen from 2 red lights in 1 way (2C2).
2 blue lights can be chosen from 5 yellow lights in 5C2 ways.
And 2 yellow lights can be chosen from 3 in 3C2 ways.
Total ways 6 lights can be chosen from 10 lights is 10C6 ways.
Answer = (90*2C2*5C2*3C2)/10C6

I know my answer for this probability question is wrong since it doesnt match any of the answer choices but can someone please guide me.

The prompt is awkward to say the least. It might be easier to think of it this way:

A jar of marbles contains two red marbles, five blue marbles, and three yellow marbles. If six marbles are drawn, one at a time, with replacement, what is the probability of drawing two of each color of marble?

Then the answer is easier to see: (1/5)² * (1/2)² * (3/10)² * 6! / (2! * 2! * 2!)

Palak-gmat wrote:I know my answer for this probability question is wrong since it doesnt match any of the answer choices but can someone please guide me.

You're assuming (more or less) that six distinct bulbs are lit at the same time. I made the same assumption when I read the problem: the wording in the prompt is junk, and I don't think the interpretation the answer requires is well supported.