Jack owns many black shirts and pants. However, Jack gets up late each day, and as a result, he just chooses his clothes at random.
He would like to have at least one black item on and he knows there is a .16 chance just his shirt will be black, there is a .27 chance just his pants will be black, and there is a chance less than both of these he won't have anything black on.
If the color of his shirt and pants are independent of each other, what is the chance both are black?
(In reply to Answer
by K Sengupta)
B = Probability that both his shirt and pants will be black.(say)
We know that:
Probability that his shirt is black but the pants are white = 16/100, and:
Probability that his shirt is white but the pants are black = 27/100
Then, we must have:
Probability that his shirt is black
= Probability that both his shirt and pants will be black + Probability that his shirt is black but the pants are white
= B + 16/100
Probability that his pants is black = B + 27/100
By the problem,
(B + 16/100) (B+27/100) = B
-> B^2 - 57*B/100 + 432/10000) = 0
-> 10000*B^2 - 5700*B + 432 = 0
-> 10000*B^2 - 4800*B - 900*B + 432 = 0
-> 100*B (100*B - 48) - 9(100*B - 48) = 0
-> (100*B - 48)(100*B - 9) = 0
-> B = 48/100, or 9/100
Now, the probability that at least one amongst the pants and the shirt is coloed black
= Prob(Black pants and white shirt) + Prob(Black pants and black shirt) + prob(black shirt and white pants)
= 27/100 + B + 16/100
= B + 43/100
-> Prob.(none of the shirt and pants is black) = 1 -(B + 43/100)
= 57/100 - B
If B = 9/100, then Prob(black shirt and white pants) = 16/100 > 9/100.
This is a contradiction.
If B = 48/100, Prob(black shirt and white pants) = 16/100 < 48/100, and:
Prob(white shirt and black pants) = 27/100 < 48/100.
This is in conformity with the given conditions.
Hence B = 48/100 = 12/25
Consequently, the required chance that both the shirt and the pants are black is 12/25.