Question

A certain carton of eggs has 3 bad eggs and 9 good eggs. An omelette is made of 3 eggs randomly chosen from the carton. What is the probability that there are
1.No bad egg
2.At least two bad eggs
3.At most one good eggs
Exactly two good eggs in the omelette

Answer 1

Answer:

A certain carton of eggs has 3 bad eggs and 9 good eggs. An omelette is made of 3 eggs randomly chosen from the carton. What is the probability that there are

1.No bad egg

2.At least two bad eggs

3.At most one good eggs

Exactly two good eggs in the omelette

Answer 2

The probability that there are

1. No bad egg = 0.3818  

2. At least two bad eggs = 0.1273

3. At most one good eggs = 0.1273

4. Exactly two good eggs in the omelet = 0.4909

Given:

A certain carton of eggs has 3 bad eggs and 9 good eggs. An omelet is made of 3 eggs    

To find:

What is the probability that there are

1.No bad egg

2. At least two bad eggs

3. At most one good eggs

4. Exactly two good eggs in the omelet    

Solution:

Formula used:

Probability = No of favorable outcomes/ Total No of outcomes

1. Probability of getting no bad egg:

No. of ways to choose 3 eggs from 12 eggs = ¹²C₃ = 220.

No. of ways that choosing 3 good eggs from 9 eggs = ⁹C₃ = 84.

The probability of getting no bad egg = 84/220 = 0.3818 (approx)

2. Probability of getting at least two bad eggs:

The number of ways to choose 2 bad eggs and 1 good egg

= ³C₂  × ⁹C₁  = 3 × 9 = 27

The number of ways to choose 3 bad eggs = ³C₃ = 1.

The probability of getting at least two bad eggs is:

= (³C₂ × ⁹C₁ +  ³C₃ /¹²C₃ = (27 + 1)/220 = 0.1273 (approx)

The probability of getting at least two bad eggs is 0.1273 (approx)

3. Probability of getting at most one good egg:

The number of ways to choose 1 good egg and 2 bad eggs

= ⁹C₁ × ³C₂ = 27.

The number of ways to choose 3 bad eggs = ³C₃  = 1

The probability of getting at most one good egg is:

= (C(9,1) * C(3,2) + C(3,3))/C(12,3) = (27 + 1)/220 = 0.1273 (approx)

The probability of getting at most one good egg is 0.1273 (approx)

4. Probability of getting exactly two good eggs:

The number of ways to choose 2 good eggs = ⁹C₂ = 36

The number of ways to choose 1 bad egg = ³C₁  = 3

The probability of getting exactly two good eggs is:

= (⁹C₂ × ³C₁)/¹²C₃  = (36 × 3)/220 = 0.4909 (approx)  

The probability of getting exactly two good eggs is 0.4909 (approx)  

Therefore,

The probability that there are

1. No bad egg = 0.3818  

2. At least two bad eggs = 0.1273

3. At most one good eggs = 0.1273

4. Exactly two good eggs in the omelet = 0.4909

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