Mathematics 9709/52 — October/November 2020

Walkthrough

We are throwing a fair die repeatedly until we get a 4. The question asks for the probability that this takes fewer than 6 throws.

First, understand what "fewer than 6 throws" means. It means the first 4 appears on throw 1, 2, 3, 4, or 5. If it takes 6 or more throws, then the first 5 throws all failed to produce a 4.

Each throw is independent, and the probability of not getting a 4 on any single throw is $\frac{5}{6}$ (since 5 of the 6 faces are not 4).

The event "6 or more throws" is the same as "no 4 in the first 5 throws". Its probability is:

$$\left(\frac{5}{6}\right)^5$$

Since "fewer than 6 throws" is the complement of "6 or more throws", we subtract from 1:

$$1 - \left(\frac{5}{6}\right)^5$$

Computing: $\left(\frac{5}{6}\right)^5 = \frac{3125}{7776}$, so:

$$1 - \frac{3125}{7776} = \frac{4651}{7776} \approx 0.598$$

This is the required probability.

Key Takeaways

• This is a geometric distribution problem: repeated independent trials until the first success.
• The complement rule is powerful: instead of summing five separate probabilities, we compute the probability of the opposite event and subtract from 1.
• Each trial has probability $\frac{1}{6}$ of success and $\frac{5}{6}$ of failure.

Common Mistakes

• Misreading "fewer than 6" as "6 or fewer", which would incorrectly include the 6th throw.
• Using $\frac{1}{6}$ as the probability of failure instead of $\frac{5}{6}$.
• Computing $\left(\frac{1}{6}\right)^5$ instead of $\left(\frac{5}{6}\right)^5$.

Things to Be Careful About

• "Fewer than 6" means $X \leq 5$, not $X \leq 6$.
• The complement event is "no 4 in the first 5 throws", not "no 4 in 6 throws".
• The answer may be given as a fraction $\frac{4651}{7776}$ or a decimal $0.598$; both are accepted by the mark scheme.

Walkthrough

This time the die is thrown exactly 10 times, and we want the probability of getting at least 3 fours.

Since there is a fixed number of throws (10) and each throw is independent with the same probability of success $\frac{1}{6}$, the number of 4s, $X$, follows a binomial distribution: $X \sim B(10, \frac{1}{6})$.

"At least 3" means $X \geq 3$. Computing $P(X \geq 3)$ directly would require summing $P(X=3) + P(X=4) + \dots + P(X=10)$, which is 8 terms. Instead, we use the complement:

$$P(X \geq 3) = 1 - P(X \leq 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$$

Now apply the binomial probability formula:

$$P(X = x) = {}^{10}C_x \left(\frac{1}{6}\right)^x \left(\frac{5}{6}\right)^{10-x}$$

For $x = 0$:

$$P(X=0) = \left(\frac{5}{6}\right)^{10} \approx 0.1615$$

For $x = 1$:

$$P(X=1) = {}^{10}C_1 \left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^9 \approx 0.3230$$

For $x = 2$:

$$P(X=2) = {}^{10}C_2 \left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^8 \approx 0.2907$$

Sum these:

$$P(X \leq 2) \approx 0.1615 + 0.3230 + 0.2907 = 0.7752$$

Then:

$$P(X \geq 3) = 1 - 0.7752 \approx 0.2248$$

Rounded to 3 decimal places, the answer is $0.225$.

Key Takeaways

• A fixed number of independent trials with constant success probability is modelled by the binomial distribution.
• The complement rule $P(X \geq k) = 1 - P(X \leq k-1)$ avoids summing many terms.
• Each binomial probability requires the combination coefficient ${}^{10}C_x$.

Common Mistakes

• Using the geometric distribution instead of the binomial distribution — part (b) has a fixed number of throws, not "until a 4 appears".
• Forgetting the combination coefficients ${}^{10}C_x$ in the binomial formula.
• Computing $P(X \geq 3)$ directly by summing 8 terms, which is more work and more error-prone.
• Rounding intermediate probabilities too early, which can change the final answer.

Things to Be Careful About

• The success probability is $\frac{1}{6}$ and the failure probability is $\frac{5}{6}$.
• The complement of "at least 3" is "0, 1 or 2", not "0, 1, 2 or 3".
• Keep at least 4 decimal places in intermediate steps so the final answer rounds correctly to $0.225$.
• The mark scheme accepts $0.2247 < p \leq 0.225$.