Mathematics 9709/52 — October/November 2021

Walkthrough

The table shows counts for male and female students for each instrument. Since every student plays exactly one instrument, the total number of students is the sum of all eight entries: 180. To find the probability that a randomly chosen student plays the guitar, add the male and female guitar players: $44 + 38 = 82$. The probability is the favourable count divided by the total: $\frac{82}{180} = \frac{41}{90}$.

Key Takeaways

A probability from a contingency table is found by dividing the number of favourable outcomes by the total number of outcomes. Fractions can be simplified, or given as decimals.

Common Mistakes

• Forgetting to add both male and female guitar players.
• Using only one row or one column instead of the relevant total.
• Using the wrong total instead of 180.

Things to Be Careful About

The question states that each student plays exactly one instrument, so the total 180 is correct. The mark scheme accepts $\frac{82}{180}$, $\frac{41}{90}$ or $0.456$.

Walkthrough

The condition 'given that the student plays the drums' restricts the sample space to drum players only. There are $11$ male drum players and $20$ female drum players, so there are $31$ drum players in total. Among these 31 students, 11 are male, so the required probability is $\frac{11}{31}$. Using the formula, $P(M \cap D) = \frac{11}{180}$ and $P(D) = \frac{31}{180}$, so $P(M|D) = \frac{11/180}{31/180} = \frac{11}{31}$.

Key Takeaways

Conditional probability restricts the sample space to the given condition. It can be computed either by counting within the restricted group or by using $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

Common Mistakes

• Using the overall probability that a student is male, $\frac{80}{180}$, instead of restricting to drum players.
• Using $\frac{11}{180}$ as the final answer without dividing by $P(D)$.
• Forgetting to include female drum players in the denominator.

Things to Be Careful About

The mark scheme accepts unsimplified fractions and condones multiplying by 180, so $\frac{11}{31}$ from the counts is acceptable. If using the formula, clearly identify $P(M \cap D)$ and $P(D)$.

Walkthrough

Let $F$ be the event 'the student is female' and $G$ be the event 'the student plays the guitar'. From the table, $P(F) = \frac{100}{180}$, $P(G) = \frac{82}{180}$ and $P(F \cap G) = \frac{38}{180}$. For independence we need $P(F \cap G) = P(F)P(G)$. Compute $P(F)P(G) = \frac{100}{180} \times \frac{82}{180} = \frac{41}{162} \approx 0.2531$. This is not equal to $\frac{38}{180} \approx 0.2111$, so the events are not independent.

Alternatively, compute $P(F|G) = \frac{P(F \cap G)}{P(G)} = \frac{38}{82} = \frac{19}{41} \approx 0.4634$. Since $P(F) = \frac{5}{9} \approx 0.5556$, these are not equal, so the events are not independent.

Key Takeaways

Two events are independent exactly when the probability of their intersection equals the product of their individual probabilities. Equivalently, $P(A|B) = P(A)$. Independence is different from mutual exclusivity.

Common Mistakes

• Concluding independence because the intersection probability is not zero.
• Comparing $P(F \cap G)$ with $P(F)$ or $P(G)$ instead of with $P(F)P(G)$.
• Forgetting to state the conclusion 'not independent' after the comparison.
• Mixing up the conditional probability method and not comparing it with the marginal probability.

Things to Be Careful About

The mark scheme requires the values and labels to be seen, for example $P(F)$, $P(G)$, $P(F \cap G)$ and $P(F) \times P(G)$. Unsimplified fractions are acceptable. The conclusion must follow from the comparison; correct values with no wrong working are sufficient.