Answers: Conditional probability
These are the answers to Questions: Conditional probability.
Please attempt the questions before reading these answers.
Q1
1.1.
- \(\mathbb{P}(A) = \dfrac{13}{52}\) (hearts)
- \(\mathbb{P}(B) = \dfrac{26}{52}\) (red cards)
- \(\mathbb{P}(A \cap B) = \dfrac{13}{52}\) (red hearts)
Using the definition of conditional probability: \[ \mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{13/52}{26/52} = \frac{1}{2} \] So the probability that the card is a heart, given that it is red, is \(1/2\).
1.2.
You are given:
- \(\mathbb{P}(\textsf{Piano} \mid \textsf{Left-handed}) = 0.25\)
So the probability that a randomly chosen student plays the piano, given that they are left-handed, is \(0.25\).
1.3.
- \(\mathbb{P}(A \cap B) = 0.15\)
- \(\mathbb{P}(B) = 0.30\)
Using the definition of conditional probability: \[ \mathbb{P}(A \mid B) = \dfrac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \dfrac{0.15}{0.30} = 0.5 \]
So the probability that an employee takes Spanish, given that they take French, is \(0.5\).
1.4.
- \(\mathbb{P}(A \cap B) = 0.25\)
- \(\mathbb{P}(B) = 0.40\)
Using the definition of conditional probability: \[ \mathbb{P}(A \mid B) = \dfrac{0.25}{0.40} = \dfrac{5}{8} = 0.625 \]
So the probability that the student is sixteen, given they bring a packed lunch, is \(0.625\).
Q2
2.1.
- \(\mathbb{P}(\textsf{first green}) = \dfrac{3}{5}\)
- \(\mathbb{P}(\textsf{second green} \mid \textsf{first green}) = \dfrac{2}{4}\)
Using the multiplication rule: \[ \mathbb{P}(\textsf{both green}) = \left(\dfrac{3}{5}\right)\left(\dfrac{2}{4}\right) = \dfrac{6}{20} = 0.3 \]
So the probability that both sweets are green is \(0.3\).
2.2.
- \(\mathbb{P}(\textsf{first pass}) = 0.9\)
- \(\mathbb{P}(\textsf{second pass} \mid \textsf{first pass}) = 0.95\)
Using the multiplication rule: \[ \mathbb{P}(\textsf{both pass}) = (0.9)(0.95) = 0.855 \]
So the probability that a box of Bayes Biscuits passes both inspections is \(0.855\).
2.3.
These are independent events, so
- \(\mathbb{P}(\textsf{heads}) = 0.5\)
- \(\mathbb{P}(\textsf{roll a 6}) = \dfrac{1}{6}\)
Using the multiplication rule: \[ \mathbb{P}(\textsf{heads and 6}) = (0.5)\left(\dfrac{1}{6}\right) = \dfrac{1}{12} \]
So the probability of getting heads and rolling a \(6\) is \(\dfrac{1}{12}\).
2.4.
- \(\mathbb{P}(\textsf{likes tea}) = 0.7\)
- \(\mathbb{P}(\textsf{likes coffee} \mid \textsf{likes tea}) = 0.6\)
Using the multiplication rule: \[ \mathbb{P}(\textsf{likes both}) = (0.7)(0.6) = 0.42 \]
So the probability that a random person likes both tea and coffee is \(0.42\).
Q3
3.1.
Given: \(\mathbb{P}(A) = 0.4\), \(\mathbb{P}(B) = 0.5\), \(\mathbb{P}(A \cap B) = 0.2\)
Check:
\[ \mathbb{P}(A)\mathbb{P}(B) = (0.4)(0.5) = 0.2 \]
Since \(\mathbb{P}(A \cap B) = \mathbb{P}(A)\mathbb{P}(B)\), the events are independent.
3.2.
Given: \(\mathbb{P}(A) = 0.3\), \(\mathbb{P}(A \mid B) = 0.3\)
Since \(\mathbb{P}(A \mid B) = \mathbb{P}(A)\), events are independent.
3.3.
Given: \(\mathbb{P}(A) = 0.5\), \(\mathbb{P}(B) = 0.4\), \(\mathbb{P}(A \cap B) = 0.1\)
Check:
\[ \mathbb{P}(A)\mathbb{P}(B) = (0.5)(0.4) = 0.2 \neq 0.1 = \mathbb{P}(A \cap B) \]
Since \(\mathbb{P}(A \cap B) \neq \mathbb{P}(A)\mathbb{P}(B)\), the events are dependent.
3.4.
Given: \(\mathbb{P}(A) = 0.6\), \(\mathbb{P}(A \mid B) = 0.2\)
Since \(\mathbb{P}(A \mid B) \neq \mathbb{P}(A)\), the events are dependent.
Version history and licensing
v1.0: initial version created 05/25 by Sophie Chowgule as part of a University of St Andrews VIP project.