If 6 bulbs are lit in a random sequence, what is the probability that no color is lit more than twice?

this is possible only if 2 bulbs of each color are lit.

now ratio is two to five to three

i.e 2/10 =red =1/5(part to whole)
5/10 = blue =1/2
3/10 = yellow

Now this is also the probabilty that bulb of each type is lit once

so if 6 bulbs are lit the reqd probabilty is

p(red is lit) X p(red is lit) X p(blue is lit) X p(blue is lit) X p(yellow is lit)
X p(yellow is lit)

=(1/5)^2 X(1/2)^2 X (3/10)^2

=9/10000

so ans should be A

samir, you need to consider the ordering also isn't ? These bulbs can be lit in 6! ways.

Hi kajcha,
Question says that 'If 6 bulbs are lit in a random sequence, what is the probability that no color is lit more than twice? "

in my soln I have considered the order
RRBBYY

now actually any of this order is possible

i.e RBRYBB, RRBYBY etc each will have probabilty of 9/10000

& total there will be 90 permutations so if we take that into consideration

then we have P(two of each is lit) = 90*9/10000 =81/1000

however this would be the Probabilty that the bulbs are lit in any Random sequence

however a Random Order means any of RBRYBB, RRBYBY ,RRBBYY etc whose individual probabilty is 9/10000

so basically we need to find the Probabilty that two of each bulbs are lit in a Random Order

so IMO ans =9/10000

Favourable outcomes for our event is that two bulbs of each color must be lit up at any place.

This is my interpretation of the question