Probability Questions
An individual is picked at random from a group of 52 athletes. Suppose that 26 of the athletes are female of which 6 are swimmers. Also, there are 10 swimmers among male athletes.
A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non-defective?
A packing plant fills bags with cement. The weight X kg of a bag of cement can be modelled by a normal distribution with mean 50kg and standard deviation 2kg.
From experience a high-jumper knows that he can clear a height of at least 1.78m once in 5 attempts. He also knows that he can clear a height of at least 1.65m on 7 out of 10 attempts. Assuming that heights the high-jumper can reach follow a Normal Distribution,
A drinks machine dispenses coffee into cups. A sign on the machine indicates that each cup contains 50ml of coffee. The machine actually dispenses a mean amount of 55ml per cup and 10% of the cups contain less than the amount stated on the sign. Assuming that the amount of coffee dispensed into each cup is normally distributed, find;
Probability Questions with Detailed Solutions
An individual is picked at random from a group of 52 athletes. Suppose that 26 of the athletes are female of which 6 are swimmers. Also, there are 10 swimmers among male athletes.
Solution:
We use conditional probability:
P(Swimmer | Female) = Number of female swimmers / Total number of females
Given:
- Total females = 26
- Female swimmers = 6
P(Swimmer | Female) = 6/26 = 3/13 ≈ 0.2308 or 23.08%
Solution:
First, find total number of swimmers:
Female swimmers = 6
Male swimmers = 10
Total swimmers = 6 + 10 = 16
Now use conditional probability:
P(Male | Swimmer) = Number of male swimmers / Total number of swimmers
P(Male | Swimmer) = 10/16 = 5/8 = 0.625 or 62.5%
A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non-defective?
Solution:
Total fruits = 10 apples + 20 oranges = 30
Case 1: Both are oranges
Number of ways to choose 2 oranges from 20:
C(20, 2) = 20!/(2! × 18!) = 190
Case 2: Both are non-defective
Non-defective apples = 10 - 3 = 7
Non-defective oranges = 20 - 5 = 15
Total non-defective = 7 + 15 = 22
Number of ways to choose 2 non-defective fruits:
C(22, 2) = 22!/(2! × 20!) = 231
Overlap: Both are oranges AND non-defective
Number of ways to choose 2 non-defective oranges:
C(15, 2) = 15!/(2! × 13!) = 105
Using the inclusion-exclusion principle:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
= C(20,2)/C(30,2) + C(22,2)/C(30,2) - C(15,2)/C(30,2)
= (190 + 231 - 105)/435 = 316/435 ≈ 0.7264 or 72.64%
A packing plant fills bags with cement. The weight X kg of a bag of cement can be modelled by a normal distribution with mean 50kg and standard deviation 2kg.
Solution:
First, calculate the z-score:
z = (x - μ)/σ = (53 - 50)/2 = 1.5
Using standard normal distribution tables:
P(X > 53) = P(Z > 1.5) = 1 - P(Z ≤ 1.5) = 1 - 0.9332 = 0.0668 or 6.68%
Solution:
We need to find x such that P(X > x) = 0.99, which means P(X ≤ x) = 0.01
Find z-score corresponding to cumulative probability 0.01:
z ≈ -2.326 (from standard normal tables)
Convert back to original scale:
x = μ + zσ = 50 + (-2.326)(2) ≈ 50 - 4.652 ≈ 45.348 kg
Therefore, 99% of bags exceed approximately 45.35 kg.
Solution:
From part (i), P(X > 53) = 0.0668
Therefore, P(X ≤ 53) = 1 - 0.0668 = 0.9332
This is a binomial probability problem with n=3, k=2, p=0.0668
P(2 >53 and 1 ≤53) = C(3,2) × (0.0668)² × (0.9332)¹
= 3 × 0.004462 × 0.9332 ≈ 0.0125 or 1.25%
From experience a high-jumper knows that he can clear a height of at least 1.78m once in 5 attempts. He also knows that he can clear a height of at least 1.65m on 7 out of 10 attempts. Assuming that heights the high-jumper can reach follow a Normal Distribution,
Solution:
[Visual sketch would show:]
- A normal distribution curve with mean μ and standard deviation σ
- 1.78m marked on the right tail with area to the right = 1/5 = 0.2
- 1.65m marked further left with area to the right = 7/10 = 0.7
Solution:
We have two equations:
1. P(X > 1.78) = 0.2 ⇒ P(X ≤ 1.78) = 0.8
Corresponding z-score ≈ 0.842
2. P(X > 1.65) = 0.7 ⇒ P(X ≤ 1.65) = 0.3
Corresponding z-score ≈ -0.524
Set up equations:
(1.78 - μ)/σ = 0.842
(1.65 - μ)/σ = -0.524
Solving these simultaneously:
From first equation: μ = 1.78 - 0.842σ
Substitute into second equation:
(1.65 - (1.78 - 0.842σ))/σ = -0.524
(-0.13 + 0.842σ)/σ = -0.524
-0.13/σ + 0.842 = -0.524
-0.13/σ = -1.366
σ ≈ 0.0952
Then μ ≈ 1.78 - 0.842(0.0952) ≈ 1.78 - 0.080 ≈ 1.700
Final answers:
Mean (μ) ≈ 1.700 m
Standard deviation (σ) ≈ 0.095 m
Solution:
Using μ = 1.700, σ = 0.095:
z = (1.74 - 1.700)/0.095 ≈ 0.421
P(X > 1.74) = P(Z > 0.421) = 1 - P(Z ≤ 0.421) ≈ 1 - 0.6635 ≈ 0.3365 or 33.65%
A drinks machine dispenses coffee into cups. A sign on the machine indicates that each cup contains 50ml of coffee. The machine actually dispenses a mean amount of 55ml per cup and 10% of the cups contain less than the amount stated on the sign. Assuming that the amount of coffee dispensed into each cup is normally distributed, find;
Solution:
Given:
- μ = 55 ml
- P(X < 50) = 0.10
Find z-score corresponding to P(Z < z) = 0.10:
z ≈ -1.2816
z = (x - μ)/σ
-1.2816 = (50 - 55)/σ
σ = -5/-1.2816 ≈ 3.901 ml
Solution:
Using μ = 55 ml, σ ≈ 3.901 ml:
z = (61 - 55)/3.901 ≈ 1.538
P(X > 61) = P(Z > 1.538) = 1 - P(Z ≤ 1.538) ≈ 1 - 0.938 ≈ 0.062 or 6.2%
Solution:
Given:
- New σ = 3 ml
- P(X < 50) = 0.025
Find z-score corresponding to P(Z < z) = 0.025:
z ≈ -1.96
z = (x - μ)/σ
-1.96 = (50 - μ)/3
μ = 50 + 1.96 × 3 = 50 + 5.88 = 55.88 ml
The new mean is approximately 55.88 ml.
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