Question

1). Given a container with 6 white balls and 9 black balls, if 4 balls are chosen randomly, what is the probability that exactly two are white?

2). Given a container with 6 white balls and 9 black balls, if 4 balls are chosen randomly, and one of the chosen balls is black, what is the probability that exactly two are white?

Answer 1

Question 1

Here in the container there are 6 white balls and 9 black balls so there are a total of 15 balls.

Now, we have to choose 4 balls , so number of possibilities of choosing 4 balls = ¹⁵C₄

Now we have to find te probability that exactly two are white that means two of them are black.

so total such possibliites = ⁶C₂⁹C₂

Probability of occuring such event = ⁶C₂⁹C₂ /¹⁵C₄ = 15 * 36/1365 = 0.3956

Question 2

Here in the container there are 6 white balls and 9 black balls so there are a total of 15 balls.

Now, we have to choose 4 balls , so number of possibilities of choosing 4 balls = ¹⁵C₄

Now we know that one of the chosen balls is black. So here there are 14 balls, out of which 6 are white and 8 are black(one we know is black), and we will choose the rest of three from it and we want to know that exactly two are white.

total possibilities for the remaining three balls = ¹⁴C₃

so total possiblities of two white balls among three= ⁶C₂ * ⁸C₁

Pr(Exactly two among four are white when one of the chosen balls is black) = ⁶C₂ * ⁸C₁/¹⁴C₃ = (15 * 8)/364 = 0.3297