Probability Distribution

Question

A bag contains 12 green and 8 orange balls that are identical except for their color.

A ball is chosen from the bag, its color noted, and then it is replaced. Eight balls are chosen.

(a) Find

(i)The probability of obtaining five green balls

(ii)The probability of obtaining at least two orange balls

(iii)The probability of obtaining equal numbers of green and orange balls.

(b) How many balls need to be chosen if the probability of obtaining at least one green ball is to be at least 99%?

I need help in part (a)'s (ii) and (iii) and in (b) part.

Answer 1

Given:

There are 20 balls in total, 12 green and 8 red.

First, get the probability of one of each color:

The probability of obtaining 1 green ball is p = 12/20 = 3/5

The probability of obtaining 1 red ball is q = 8/20 = 2/5.

(WATCH THE PATTERN!)

The probability distribution for obtaining 8 balls, where _nC_r = n!/(r!(n-r)!), is:

(1) For exactly 8 green balls:

₈C₀ • p⁸ = (3/5)⁸ ≈ 0.01679616

(2) For 7 green balls and 1 red ball:

₈C₁ • p⁷q¹ = 8(3/5)⁷(2/5)¹ ≈ 0.08957952

(3) For 6 green balls and 2 red balls:

₈C₂ • p⁶q² = 28•(3/5)⁶(2/5)² ≈ 0.20901888

(4) For 5 green balls and 3 red balls:

₈C₃ • p⁵q³ = 56•(3/5)⁵(2/5)³ ≈ 0.27869184

(5) For 4 green balls and 4 red balls:

₈C₄ • p⁴q⁴ = 70•(3/5)⁴(2/5)⁴ ≈ 0.2322432

(6) For 3 green balls and 5 red balls:

₈C₅ • p³q⁵ = 56•(3/5)³(2/5)⁵ ≈ 0.12386304

(7) For 2 green balls and 6 red balls:

₈C₆ • p²q⁶ = 28•(3/5)²(2/5)⁶ ≈ 0.04128768

(8) For 1 green balls and 7 red balls:

₈C₇ • p¹q⁷ = 8•(3/5)¹(2/5)⁷ ≈ 0.00786432

(9) For exactly 8 red balls:

₈C₈ • q⁸ = 1•(2/5)⁸ ≈ 0.00065536

(a)

(i) The probability of obtaining five green balls:

₈C₃ • p⁵q³ = 56•(3/5)⁵(2/5)³ ≈ 0.27869184

(ii)The probability of obtaining at least two orange balls:

There is no orange. Therefore the probability is 0

(iii)The probability of obtaining equal numbers of green and orange balls:

₈C₄ • p⁴q⁴ = 70•(3/5)⁴(2/5)⁴ ≈ 0.2322432

(b) How many balls need to be chosen if the probability of obtaining at least one green ball is to be at least 99%?

"At least one green ball" means all of the above except "For exactly all red balls". Therefore we need to subtract (2/5)ⁿ, where n is the number of balls chosen, from 100% ( or 1).

For 5 balls:

1 - (2/5)⁵ ≈ 0.98976, which is less than 99%

For 6 balls:

1 - (2/5)⁶ ≈ 0.995904, which is greater than 99%

Therefore at least 6 balls needed to be chosen if the probability of obtaining at least one green ball is to be at least 99%. If you choose 5 balls or below, you'll get less than 99%.