Mathematics 9709/52 — May/June 2021

Walkthrough

The die is fair, so on any single throw the probability of getting a 5 is $\frac{1}{6}$. We keep throwing until the first 5 appears, so $X$ counts the number of trials up to and including that first 5. This is exactly the geometric distribution with success probability $p=\frac{1}{6}$.

For a geometric distribution, the expected number of trials needed for the first success is $\frac{1}{p}$. In this case:

$$\text{E}(X) = \frac{1}{1/6} = 6$$

So on average, it takes 6 throws to obtain the first 5.

Key Takeaways

A situation of repeated independent trials until the first success is modelled by a geometric distribution. The parameter $p$ is the probability of success on each trial, and its mean is $\frac{1}{p}$. Recognising this structure immediately avoids lengthy expectation calculations.

Common Mistakes

A common mistake is to take the mean as $np$, the formula for the binomial distribution. Here there is no fixed number of trials, so the geometric mean $\frac{1}{p}$ is correct.

Another common mistake is to use $p=\frac{5}{6}$ because there are five non-5 outcomes. The success being counted is obtaining a 5, so $p=\frac{1}{6}$.

Things to Be Careful About

The distribution counts the total number of throws including the throw on which the 5 first appears, not the number of failures before it. The formula $\frac{1}{p}$ is for this total-trial count.

Walkthrough

For a geometric distribution, if $X=k$ then the first $k-1$ throws are not 5 and the $k$th throw is a 5. With $p=\frac{1}{6}$ and $q=\frac{5}{6}$,

$$\text{P}(X=k)=q^{k-1}p.$$

The phrase after the 3rd throw but before the 8th throw excludes the 3rd throw itself and the 8th throw itself, so it includes the 4th, 5th, 6th and 7th throws. We therefore add four separate probabilities. Since these events are mutually exclusive, addition is valid.

To simplify the sum, factor out $q^3p$. The remaining bracket is a finite geometric sum, and multiplying by $1-q$ gives $1-q^4$. This produces

$$q^3 - q^7.$$

Equivalently, this is $\text{P}(X\le 7)-\text{P}(X\le 3)=q^3-q^7$. Finally substitute $q=\frac{5}{6}$ and evaluate to 0.2996..., which is 0.300 to three significant figures.

Key Takeaways

The geometric distribution gives the probability that the first success occurs on the $k$th trial. Summing probabilities over a range is appropriate when multiple outcomes satisfy the condition, because these outcomes are mutually exclusive. Recognising a common factor leads to a compact expression.

Common Mistakes

A common mistake is to include $X=3$ or $X=8$ by misreading after the 3rd but before the 8th. The phrase means $X$ takes values strictly between 3 and 8.

Another frequent error is using the wrong exponent: the probability for $X=k$ is $q^{k-1}p$, not $q^{k}p$.

Things to Be Careful About

The die is fair, so $p=\frac{1}{6}$ and $q=\frac{5}{6}$. Give the final probability to at least 3 significant figures, as requested by the mark scheme. The exact value is 0.2996..., so 0.300 is correct to 3 s.f.

Walkthrough

Here we want $\text{P}(X<10)$. Since $X$ is the number of throws until the first 5, $X<10$ means the first 5 appears on one of throws 1 to 9. The complement is $X\ge 10$, which happens exactly when the first 9 throws all fail to give a 5.

The probability of no 5 on one throw is $q=\frac{5}{6}$. Therefore

$$\text{P}(X\ge 10)=q^9=\left(\frac{5}{6}\right)^9.$$

Using the complement rule,

$$\text{P}(X<10)=1-q^9=0.80619\ldots$$

An equivalent method is to add the geometric probabilities for $X=1,2,\ldots,9$:

$$\frac16+\frac16\left(\frac56\right)+\cdots+\frac16\left(\frac56\right)^8=0.806.$$

Key Takeaways

For a waiting-time geometric distribution, the complement of first success within $n$ trials is no success in the first $n$ trials. This often gives the fastest calculation.

Common Mistakes

A common mistake is to use $1-q^{10}$ because fewer than 10 is confused with at most 10. Fewer than 10 throws means $X\le 9$, so the complement is $q^9$, not $q^{10}$.

Another possible mistake is to set $q=\frac16$ instead of $\frac56$ when considering failures.

Things to Be Careful About

The wording fewer than 10 is strict: it includes 9 but not 10. Use the complement $q^9$. The answer should be given as 0.806 to 3 significant figures.