A bag contains 8 red and 5 white balls. Two successive draws of 3 balls are made without replacement. The probability that the first draw will produce 3 white balls and the second 3 red balls is

• (A) 5223
• (B) 6257
• (C) 7429
• (D) 3548

To find the probability of drawing 3 white balls first and then 3 red balls from a bag containing 8 red and 5 white balls, we can break the problem down into two parts: the first draw and the second draw.

Calculating the Probability of the First Draw

For the first draw, we need to calculate the probability of drawing 3 white balls from the 5 available. The total number of ways to choose 3 balls from 13 (8 red + 5 white) is given by:

• Total ways to choose 3 balls: C(13, 3) = 286
• Ways to choose 3 white balls: C(5, 3) = 10

The probability of drawing 3 white balls is:

P(3 white) = C(5, 3) / C(13, 3) = 10 / 286

Calculating the Probability of the Second Draw

After drawing 3 white balls, there are now 10 balls left (8 red and 2 white). We now need to find the probability of drawing 3 red balls:

• Total ways to choose 3 balls from 10: C(10, 3) = 120
• Ways to choose 3 red balls: C(8, 3) = 56

The probability of drawing 3 red balls is:

P(3 red | 3 white drawn) = C(8, 3) / C(10, 3) = 56 / 120

Combining the Probabilities

The overall probability of both events happening (drawing 3 white balls first and then 3 red balls) is the product of the two probabilities:

P(total) = P(3 white) * P(3 red | 3 white drawn)

Calculating this gives:

P(total) = (10 / 286) * (56 / 120) = 560 / 34320

Simplifying this fraction results in:

P(total) = 1 / 61.5

Final Answer

After evaluating the options provided, the probability that the first draw will produce 3 white balls and the second draw will produce 3 red balls is approximately:

(B) 6257