[email protected] wrote:Hi Palak-gmat,

This question was discussed here:

https://www.beatthegmat.com/a-string-of- ... 15774.html

GMAT assassins aren't born, they're made,
Rich

Palak-gmat 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? 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.

A) 9/10000
B) 81/1000
C) 10/81
D) 1/3
E) 80/81

Answer:B

Matt@VeritasPrep wrote:

Palak-gmat 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? 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.

A) 9/10000
B) 81/1000
C) 10/81
D) 1/3
E) 80/81

Answer:B

I think you were confused (as it was) by the clumsiness of the prompt. It seems to assume that the lights are lit one by one, so that the same light might be lit more than once. For instance, when we light the first bulb, its chances of being red are 1/5. Then, the bulb goes out, and we light another bulb at random ... and it could be the same one again!

If we assume that the lights must all be lit at the same time, your approach is on the right track ... but the problem seems not to be doing that. I don't know how anyone would deduce that from the prompt, however, without guessing: "6 bulbs are lit" makes me immediately think of six distinct bulbs.

Palak-gmat wrote:Thanks a ton I now understand where I was going wrong.