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02 Oct 2018, 02:30
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C
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A box contains 16 balls, of which 4 are white, 3 are blue, and the rest are yellow. If two balls are to be selected at random from the box, one at a time without being replaced, what is the probability that one ball selected will be white and the other ball selected will be blue?
(A) 5/64
(B) 1/16
(C) 1/10
(D) 1/5
(E) 1/6
(A) 5/64
(B) 1/16
(C) 1/10
(D) 1/5
(E) 1/6
02 Oct 2018, 04:17
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We have 16 balls , of which 4 are white , 3 are blue, and the rest that is 16-7 = 9 balls are yellow.
We are drawing 2 balls, and we need to find the probability of them being one white and one blue.
This can happen in two ways.
First way is drawing a white ball first, and the blue ball second.
So, the probability of drawing the white ball first is = 4/16 = 1/4
The probability of drawing the blue ball second is = 3/15 = 1/5
So, multiplying the both we get, = (1/4)x(1/5) = 1/20 (let this be way 1)
Second way is drawing the blue ball first and drawing the white ball second.
So, the probability of drawing the blue ball first is = 3/16
The probability of drawing the white ball second is = 4/15
Multiplying the above both we get , = (3/16)x(4/15) = 1/20 (let this be way 2)
Adding both ways we get, = (1/20)+(1/20) = 2/20 = 1/10.
So the correct answer is C.
We are drawing 2 balls, and we need to find the probability of them being one white and one blue.
This can happen in two ways.
First way is drawing a white ball first, and the blue ball second.
So, the probability of drawing the white ball first is = 4/16 = 1/4
The probability of drawing the blue ball second is = 3/15 = 1/5
So, multiplying the both we get, = (1/4)x(1/5) = 1/20 (let this be way 1)
Second way is drawing the blue ball first and drawing the white ball second.
So, the probability of drawing the blue ball first is = 3/16
The probability of drawing the white ball second is = 4/15
Multiplying the above both we get , = (3/16)x(4/15) = 1/20 (let this be way 2)
Adding both ways we get, = (1/20)+(1/20) = 2/20 = 1/10.
So the correct answer is C.
