Question 9
A basket contains 10 green apples and 30 red apples. Three apples are drawn at random from the basket with replacement. Determine the probability that exactly two of the three apples are green.
Not quite. $\tfrac{3}{64}=\left(\tfrac14\right)^2\left(\tfrac34\right)$ is one path, such as green, green, red. The red apple can come first, second or third, so there are three paths. Multiply by $\binom32=3$. In 2021, 27.5% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2021, 50.7% of Queensland students chose (B).
Not quite. $\tfrac{10}{64}=\tfrac{9}{64}+\tfrac{1}{64}$ is at least two green, which also counts all three green. The question says exactly two, so leave out $\tfrac{1}{64}$. In 2021, 9.5% of Queensland students chose (C).
Not quite. $\tfrac{27}{64}=3\left(\tfrac14\right)\left(\tfrac34\right)^2$ is exactly one green, which is the same as exactly two red. Only 10 of the 40 apples are green, so a green apple has probability $\tfrac14$. In 2021, 11.5% of Queensland students chose (D).
G is a green apple, probability $\tfrac{10}{40}=\tfrac14$. R is red, probability $\tfrac34$. Each apple goes back, so every draw has the same probabilities.
Three paths have exactly two green apples: GGR, GRG and RGG. Each has probability $\tfrac{3}{64}$, which is option (A) on its own. Together they give $\tfrac{9}{64}$. That’s option (B).
At least two also counts GGG, which adds $\tfrac{1}{64}$. That gives $\tfrac{10}{64}$, option (C), but the question asks for exactly two.
Exactly one green apple is the same event as exactly two red: three paths of $\tfrac{9}{64}$, total $\tfrac{27}{64}$. That’s option (D), but the question asks for two green.
No green apples is just RRR, $\tfrac{27}{64}$. That happens to equal exactly one green, also $\tfrac{27}{64}$, but neither is what the question asks.
All three green is just GGG, $\tfrac{1}{64}$. The question asks for exactly two.
At least three green can only be GGG, so it is the same as exactly three, $\tfrac{1}{64}$.
At least one green is every path except RRR: $1-\tfrac{27}{64}=\tfrac{37}{64}$.
Every path has at least zero green apples, so all eight count and the total is 1.
10 of the 40 apples are green, so one draw is green with probability $\tfrac{10}{40}=\tfrac14$. Each apple goes back before the next draw, so that probability stays the same every time.
That makes the number of green apples binomial, with $n=3$ and $p=\tfrac14$.
One way to get two green is green, green, red.
$\dfrac14\times\dfrac14\times\dfrac34=\dfrac{3}{64}$
The red apple could be first, second or third. That’s $\binom32=3$ paths, each with the same probability.
Add the three paths, or use the binomial formula.
$P(X=2)=\dbinom32\left(\dfrac14\right)^2\left(\dfrac34\right)$$=3\times\dfrac{3}{64}=\dfrac{9}{64}$
$$\begin{aligned}P(X=2)&=\binom32\left(\frac14\right)^2\left(\frac34\right)^1\\&=3\times\frac{3}{64}\\&=\frac{9}{64}\end{aligned}$$
With replacement, each draw is green with probability $\tfrac14$, and there are three orders for two green and one red.
How Queensland students answered in 2021
From the QCAA Mathematical Methods subject report for 2021. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2021 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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