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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