GATE ME 2012 | Question: 45

Question

A box contains $4$ red balls and $6$ black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is

• A. $1/20$
• B. $1/12$
• C. $3/10$
• D. $1/2$

Answer 1

Given, Red balls = 4 and Black balls = 6

So, Total balls = 10

Now, there are three ways of getting 1 red and 2 black balls: RBB or BRB or BBR. Using this information we can simply solve for selection without replacement as follows:

 P({R, B, B}) = (4/10) * (6/9) * (5/8) = 1/6

or P({B, R, B}) =  (6/10) * (4/9) * (5/8) = 1/6

or P({B, B, R}) = (6/10) * (5/9) * (4/8) = 1/6

Therefore, P(RBB or BRB or BBR) = (1/6) + (1/6) + (1/6) = 3/6 = 0.50

Hence, Option D is correct.