Question

Arthur has a box which contains 6 black balls and n (n ≥ 2) white balls, then at random, he takes 3 balls from the box. If the probability of taking any ball is the same and Pₙ is the probability of taking 1 black ball and 2 white balls, what is the largest possible value for Pₙ?​

Answer 1

• 11/68 or 0.1618

Explanation:

Balls

• Black - 6
• White - n

Options of getting 3 balls

• 1. White, black, black
• 2. Black, white, black
• 3. Black, black, white

Probability P(n) in each option

1. WBB

• P(w) = 6/(n + 6)
• P(b1) = n/(n + 6 - 1) = n/(n + 5)
• P(b2) = (n - 1)/(n + 5 - 1) = (n - 1)/(n + 1)

P(n) =

• P(w)P(b1)(P(b2) =
• 6/(n+6) × n/(n + 5) × (n - 1)/(n + 4) =
• 6n(n - 1)/(n + 6)(n + 5)(n + 4)

2. BWB

• P(b1) = n/(n + 6)
• P(w) = 6/(n + 6 - 1) = 6/(n + 5)
• P(b2) = (n - 1)/(n + 5 - 1) = (n - 1)/(n + 4)

P(n) =

• P(b1)P(w)(P(b2) =
• n/(n+6) × 6/(n + 5) × (n - 1)/(n + 4) =
• 6n(n - 1)/(n + 6)(n + 5)(n + 4)

3. BBW

• P(b1) = n/(n + 6)
• P(b2) = (n - 1)/(n + 6 - 1) = (n - 1)/(n + 5)
• P(w) = 6/(n + 5 - 1) = 6/(n + 4)

P(n) =

• P(b1)P(b2)(P(w) =
• n/(n+6) × (n - 1)/(n + 5) × 6/(n + 4) =
• 6n(n - 1)/(n + 6)(n + 5)(n + 4)

Final equation is same for each case:

• P(n) = 6n(n - 1) / (n + 6)(n + 5)(n + 4)

The easy way to find the maximum is to try the numbers or graph.

Both of the methods give the maximum integer n = 11 or n = 12

See attached graph

At both values n we get P(n):

• P(11) = 6*11*10 / 15*16*17 = 11/68 = 0.1618 (rounded)
• P(12) = 6*12*11 / 16*17*18 = 11/68 = 0.1618 (rounded)