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?
9/10000
81/1000
10/81
1/3
80/81
Probability
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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
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
Regards
Samir
Samir
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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
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
Regards
Samir
Samir