Question 17
In any five-day working week Leonardo either catches a bus to work or uses another form of transportation. On average, he catches the bus to work on three of the five days. His decision on any given day is independent of his decision on any other day.
Determine the probability that Leonardo catches a bus to work on exactly one day in a given five-day working week.
[3 marks]Five days, five trials, one bus day
Watch an average of three bus days a week become the probability for each day. Then count the ways to catch the bus on only one day.
The average gives you $p=\frac{3}{5}$. The five rows are the $\binom{5}{1}=5$ ways to choose the one bus day.
Find the $p$ that gives three bus days on average
The question never tells you $p$. It tells you the mean. Slide $p$ until the mean of the binomial distribution is 3 days. The orange bar is the probability you want.
The mean is $5\times\frac{3}{5}=3$ bus days a week. That is what “on average, he catches the bus to work on three of the five days” says, so this is the $p$ for the question.
With $p=\frac{1}{5}$ the mean is only 1 bus day a week. The “one day” in the question is the number of successes you want. It belongs in $P(X=1)$, and $p$ comes from the average.
The mean is bus days a week, fewer than three. Move $p$ up.
The mean is bus days a week, more than three. Move $p$ down.
Pick a day. The other four days are then all “other”.
This week has one factor of $p$ and four factors of $1-p$, whichever day you pick. You have found of the arrangements. Try another day.
That is every way to have exactly one bus day. There are $\binom{5}{1}=5$ of them, each with the same probability, so you multiply by 5.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The QCAA’s sample response gets $n$ and $p$ first, then writes the binomial expression, then simplifies it by hand.
Each day has two outcomes, the days are independent and the chance of the bus is the same every day. That makes this a binomial distribution. Let $X$ be the number of days he catches the bus. Five days means five trials. “On average” tells you that three bus days is the mean, and the mean of a binomial is $np$.
$$\begin{aligned}n&=5\\np&=3\\p&=\frac{3}{5}\\q&=1-p=\frac{2}{5}\end{aligned}$$
Marker: correctly determines the values for $p$ and $n$
“Exactly one day” is $P(X=1)$. Put $r=1$ into the binomial formula. The $\binom{5}{1}$ counts the five possible days for the one bus trip.
$$\begin{aligned}P(X=1)&=\binom{5}{1}\left(\frac{3}{5}\right)^1\\&\quad\times\left(\frac{2}{5}\right)^4\end{aligned}$$
Equivalent forms are accepted, such as $npq^{n-1}$. If your $n$ or $p$ was wrong in step 1, you can still earn this mark by follow-through.
Marker: establishes expression for the required probability
This is technology-free, so simplify by hand. Cancel the 5 from $\binom{5}{1}=5$ against the denominator of $\frac{3}{5}$ before you multiply.
$$\begin{aligned}P(X=1)&=5\times\frac{3}{5}\times\frac{2^4}{5^4}\\&=3\times\frac{16}{5^4}\\&=\frac{48}{625}\end{aligned}$$
Follow-through applies here as well if an earlier $n$ or $p$ was wrong.
Marker: determines probability
Each phrase of the question gave you something
A simple familiar binomial question hands you $n$ and $p$. This one hands you a mean, three of the five days “on average”, and you have to turn it into $p$ through $np=3$. You also have to recognise the binomial setting from the wording and then carry the exact fractions through by hand. It stays familiar because the question names the two outcomes and says the days are independent, which points straight at the binomial model.
If I had to guess where you might lose marks, it would be the first one. Reading “one day” or “three days” as the probability gives $p=\frac{1}{5}$ or nonsense like $p=3$. The next risk is dropping the $\binom{5}{1}$ and giving $\frac{48}{3125}$, which is the chance of one particular arrangement, such as a bus on Monday only. Then there is the arithmetic. The 2021 subject report observes that students find manipulating fractions by hand challenging, so cancel the 5 early and keep powers of 5 until the last line.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2021 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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