Question

Consider an urn containing 8 white balls, 7 red balls and 5 black balls.

Answer 1

1. When 2 balls are randomly selected without replacement, the probability of getting two (2) white balls is 0.1474.
2. When 5 balls are randomly selected without replacement, the probability of getting two (2) white balls is 0.3456.
3. When 150 balls are randomly selected with replacement, the probability of getting at least seventy two (72) white balls is 0.7948.

How to determine the probabilities?

First of all, we would determine the total number of balls in the urn as follows:

Total number of balls = 8 + 7 + 5

Total number of balls = 20 balls.

Next, we would determine the probability of getting two (2) white balls without replacement:

P(2 white balls) = 8/20 × 7/19

P(2 white balls) = 2/5 × 7/19

P(2 white balls) = 0.1474.

Part 2.

When 5 balls are selected without replacement, the probability of getting two (2) white balls would be calculated as follows:

P = [⁵C₂ × (8/20 × 7/19) × (12/18 × 11/17 × 10/16)]

P = [5!/(2! × (5 - 2)!) × (2/5 × 7/19) × (2/3 × 11/17 × 5/4)]

P = [5!/(2! × 3!) × (2/5 × 7/19) × (2/3 × 11/17 × 5/4)]

P = [20/2 × (2/5 × 7/19) × (2/3 × 11/17 × 5/4)]

P = [10 × (2/5 × 7/19) × (2/3 × 11/17 × 5/4)]

P = 0.3456.

Part 3.

When 150 balls are randomly selected with replacement, the probability of getting at least seventy two (72) white balls would be calculated by applying binomial probability equation. Mathematically, binomial probability is given by this equation:

$P =\; ^nC_r (p)^r (q)^{(n-r)}$

Substituting the given parameters into the formula, we have;

P = [¹⁵⁰C₇₂ × (8/20)⁷² × (8/20)⁽¹⁵⁰ ⁻ ⁷²⁾]

P = [150!/(72! × (150 - 72)!) × (8/20)⁷² × (8/20)⁽¹⁵⁰ ⁻ ⁷²⁾]

P = [150!/(72! × (78)!) × (4/5)⁷² × (4/5)⁽⁷⁸⁾]

P = 0.7948.

Read more on probability here: brainly.com/question/14805135

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Answer 2

Answer + Step-by-step explanation:

1) The probability of getting 2 white balls is equal to:

$=\frac{8}{20} \times \frac{7}{19}\\\\= 0.147368421053$

2) the probability of getting 2 white balls is equal to:

$=C^{2}_{5}\times (\frac{8}{20} \times \frac{7}{19}) \times (\frac{12}{18} \times \frac{11}{17} \times \frac{10}{16})\\=0.397316821465$

3) The probability of getting at least 72 white balls is:

$=C^{72}_{150}\times \left( \frac{8}{20} \right)^{72} \times \left( \frac{7}{20} \right)^{78} +C^{73}_{150}\times \left( \frac{8}{20} \right)^{73} \times \left( \frac{7}{20} \right)^{77} + \cdots +C^{149}_{150}\times \left( \frac{8}{20} \right)^{149} \times \left( \frac{7}{20} \right)^{1} +\left( \frac{8}{20} \right)^{150}$

$=\sum^{150}_{k=72} [C^{k}_{150}\times \left( \frac{8}{15} \right)^{k} \times \left( \frac{7}{15} \right)^{150-k}]$