Question

Three boxes contain 6 red, 4 black; 5 red, 5 black and 4 red, 6 black balls respectively. One of the box is selected at random and a ball is drawn from it. If the ball drawn is red, then the probability that it is drawn from the first bag is k. The value of 10k is___.

Answer 1

The probability of drawing a red ball from the first box is 3/5. The value of 10k is 3/50.

Step-by-step explanation:

To find the probability that a red ball is drawn from the first box, we need to calculate the probability of selecting the first box given that a red ball is drawn.

Let's denote the events as follows:

A: Selecting the first box

R: Drawing a red ball

The probability of selecting the first box given that a red ball is drawn, denoted as P(A|R), can be calculated using Bayes' theorem:

P(A|R) = (P(R|A) * P(A)) / P(R)

In this case, P(R|A) is the probability of drawing a red ball from the first box, which is 6/10. P(A) is the probability of selecting the first box, which is 1/3. And P(R) is the probability of drawing a red ball, which can be calculated as:

P(R) = (P(R|A) * P(A)) + (P(R|B) * P(B)) + (P(R|C) * P(C))

where P(B) and P(C) are the probabilities of selecting the second and third boxes, respectively.

Using the given information:

P(R) = (6/10 * 1/3) + (5/10 * 1/3) + (4/10 * 1/3)

P(R) = 1/2

Substituting the values into the formula:

P(A|R) = (6/10 * 1/3) / (1/2)

P(A|R) = 12/20 = 3/5

Therefore, 10k = 3/5 => k = (3/5) / 10 = 3/50

The value of 10k is 3/50.