How many QCAA Mathematical Methods questions cover Discrete random variables?

AusGrader has 131 QCAA Mathematical Methods questions on Discrete random variables, all with instant AI grading and detailed marking feedback.

Complete the probability distribution for $X$ in the table below. | $x$ | 0 | 1 | 2 | |---|---|---|---| | $\text{P}(X = x)$ | $\frac{3}{10}$ | | |

| $x$ | 0 | 1 | 2 | |---|---|---|---| | $\text{P}(X=x)$ | $\frac{3}{10}$ | $\frac{2}{5} \times \frac{3}{4} + \frac{3}{5} \times \frac{1}{2} = \frac{3}{5}$ | $\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$ |

Complete the table below by giving the probability of each outcome. | $x$ | 0 | 1 | 2 | 3 | 4 | |---|---|---|---|---|---| | $\text{Pr}(X = x)$ | $\frac{1}{16}$ | | $\frac{6}{16}$ | | |

| $x$ | 0 | 1 | 2 | 3 | 4 | |---|---|---|---|---|---| | $\text{Pr}(X = x)$ | $\frac{1}{16}$ | $\frac{4}{16}$ | $\frac{6}{16}$ | $\frac{4}{16}$ | $\frac{1}{16}$ |

QCAA Mathematical Methods — Discrete random variables Questions & Answers

Q12

2022

QCAA

Paper 1

3 marks

Q12

The probability that a debating team wins a debate can be modelled as a Bernoulli distribution. Given that the probability of winning a debate is $\frac{4}{5}$

Q12a

1 mark

Determine the mean of this distribution.

Reveal Answer

$x_i$	0	1
$p_i$	$\frac{1}{5}$	$\frac{4}{5}$

$E(X) = \sum p_i x_i$
$= \frac{1}{5} \times 0 + \frac{4}{5} \times 1$
$= \frac{4}{5}$

Marking Criteria

Descriptor	Marks
correctly determines the mean	1

Q12b

1 mark

Determine the variance of this distribution.

Reveal Answer

$Var(X) = \sum p_i (x_i - \mu)^2$
$= \frac{1}{5} \times \left(\frac{-4}{5}\right)^2 + \frac{4}{5} \times \left(\frac{1}{5}\right)^2$
$= \frac{4}{25}$

Marking Criteria

Descriptor	Marks
correctly determines the variance	1

Q12c

1 mark

Determine the standard deviation of this distribution.

Reveal Answer

Standard deviation $= \sqrt{Variance}$
$= \frac{2}{5}$

Marking Criteria

Descriptor	Marks
determines the standard deviation	1

Q6

2020

QCAA

Paper 1

1 mark

Q6

1 mark

If the probability of success in a Bernoulli trial is 0.30, the variance is

A

0.70

B

0.46

C

0.30

D

0.21

Reveal Answer

A

0.70

This value represents the probability of failure ( $q = 1 - p = 0.70$ ), not the variance.

B

0.46

This value is approximately the standard deviation ( $\\sqrt{0.21} \\approx 0.46$ ), rather than the variance.

C

0.30

This is the probability of success ( $p$ ), not the variance.

D

0.21

Correct Answer

The variance of a Bernoulli trial is calculated using the formula $p(1-p)$ . With $p=0.30$ , the variance is $0.30 \\times 0.70 = 0.21$ .

Q17

2025

QCAA

Paper 1

6 marks

Q17

6 marks

In a computer fishing game, a player repeatedly casts their hook into either a blue pond or a red pond. Each fish they catch scores points.

In the blue pond, the probability of catching a fish on a cast is $\frac{2}{3}$ and each fish caught scores 10 points.

In the red pond, the probability of catching a fish on a cast is $\frac{1}{3}$ and each fish caught scores 15 points.

A player has three casts left and needs to score at least 30 points to win. All remaining casts must be in the same pond.

It is claimed that the probability of winning if casting in the blue pond is $\frac{1}{27}$ more than the probability of winning if casting in the red pond. Evaluate the reasonableness of the claim.

Reveal Answer

Let $X$ be a binomial random variable representing the number of casts that caught a fish.
$p$ corresponds to a probability of a catch (success).

Blue pond:
$X \sim B\left(3, \frac{2}{3}\right)$
Red pond:
$X \sim B\left(3, \frac{1}{3}\right)$

In the blue pond, 30 points will require three successes in three casts:

P(X = 3) = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \left(\frac{2}{3}\right)^3 = \frac{8}{27}

In the red pond, 30 points will require at least two successes in three casts:
i.e. $X \ge 2$

\begin{align*} P(X \ge 2)&= P(X = 2) + P(X = 3)\\ &= \binom{3}{2} p^2 q^1 + \binom{3}{3} p^3 q^0\\ &= 3\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right) + 1\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^0\\ &= \frac{6}{27} + \frac{1}{27}\\ &= \frac{7}{27} \end{align*}

In the blue pond, the required probability is $\frac{8}{27}$
In red pond, the required probability is $\frac{7}{27}$

\frac{8}{27} - \frac{7}{27} = \frac{1}{27}

The claim is correct. Probability of winning if casting in the blue pond is $\frac{1}{27}$ more than the probability of winning using the red pond.

Marking Criteria

Descriptor	Marks
correctly identifies a method to determine at least 30 points in three casts for the blue pond	1
determines the probability of scoring at least 30 points in the blue pond	1
correctly identifies a method to determine at least 30 points in three casts for the red pond	1
substitutes appropriate values into the chosen method for the red pond	1
determines the probability of scoring at least 30 points in the red pond	1
evaluates the claim	1

Q9

2020

SCSA

Paper 2

8 marks

Q9

A cake shop makes birthday cakes. The probability distribution of the number of birthday cakes sold in a day, $X$ , is given below.

$x$	0	1	2	3	4
$P(X = x)$	0.1	0.2	0.25	0.35	0.1

Q9c

On Tuesday, the shop makes three birthday cakes. Let the random variable $Y$ denote the number of birthday cakes not sold on that day.

Q9a

1 mark

Calculate the mean number or birthday cakes sold in a day.

Reveal Answer

$E(X) = 0 \times 0.1 + 1 \times 0.2 + 2 \times 0.25 + 3 \times 0.35 + 4 \times 0.1 = 2.15$

Marking Criteria

Descriptor	Marks
evaluates the mean correctly	1

Q9b

3 marks

On Monday, the cake shop makes four birthday cakes. If each birthday cake costs ＄20 to make and sells for ＄50, what is the expected profit or loss on that day?

Reveal Answer

The cost of making four birthday cakes is $4 \times ＄20 = ＄80$
The expected number sold is $E(X) = 2.15$ , so the expected income is
$2.15 \times ＄50 = ＄107.50$
Thus, the expected profit for the shop is $＄107.50 - ＄80 = ＄27.50$

Marking Criteria

Descriptor	Marks
calculates the cost of making the birthday cakes	1
uses the expected number sold to calculate the expected income	1
calculates the expected profit	1

Q9c

2 marks

Explain why $P(Y = 0) = 0.45$ .

Reveal Answer

$Y = 0$ when all the birthday cakes are sold
So, the number of birthday cakes requested for sale is 3 or 4
So $P(Y = 0) = P(X = 3) + P(X = 4) = 0.35 + 0.1 = 0.45$

Marking Criteria

Descriptor	Marks
relates the value of $Y$ to the values of $X$	1
obtains the probability as a sum of the two probabilities	1

Q9d

2 marks

Obtain the probability distribution of $Y$ .

Reveal Answer

$y$	0	1	2	3
$P(Y = y)$	0.45	0.25	0.2	0.1

Marking Criteria

Descriptor	Marks
obtains $P(Y = 1)$ or $P(Y = 2)$ or $P(Y = 3)$	1
completes table correctly	1

Q15

2023

QCAA

Paper 1

4 marks

Q15

In a certain game, players throw one water balloon at a target. There is a one in four chance of hitting the target.

Q15a

2 marks

State the probabilities of all the possible outcomes for one throw at the target.

Reveal Answer

There are two outcomes: hit or miss.
Hit $P(\text{hit}) = \frac{1}{4}$
Miss $P(\text{miss}) = \frac{3}{4}$

Marking Criteria

Descriptor	Marks
Correctly states the probability of a 'hit'	1
Correctly states the probability of a 'miss'	1

Q15b

2 marks

Let $H$ be the discrete random variable for one of the possible outcomes. Determine the mean and variance of the distribution of random variable $H$ when 20 players throw a water balloon at the target.

Reveal Answer

Using $H$ to represent the number of 'hits'.
The mean of a binomial distribution is $np$ .
$np = 20 \times \frac{1}{4} = 5$
The variance of a binomial distribution is $np(1-p)$ .
$np(1-p) = 5 \times \frac{3}{4}$
$= \frac{15}{4} = 3\frac{3}{4}$

Marking Criteria

Descriptor	Marks
Determines the mean	1
Determines the variance	1

Q5

2020

VCAA

Paper 1

4 marks

Q5

For a certain population the probability of a person being born with the specific gene SPGE1 is $\frac{3}{5}$ .

The probability of a person having this gene is independent of any other person in the population having this gene.

Q5a

2 marks

In a randomly selected group of four people, what is the probability that three or more people have the SPGE1 gene?

Reveal Answer

$X \sim \text{Bi}\left(4, \frac{3}{5}\right)$

$\Pr(X \geq 3)$

$= \Pr(X = 3) + \Pr(X = 4)$

$= \binom{4}{3}\left(\frac{3}{5}\right)^3\left(\frac{2}{5}\right) + \left(\frac{3}{5}\right)^4$

$= \frac{297}{625}$

Marking Criteria

Descriptor	Marks
Formulates the correct probability expression for $\Pr(X \geq 3)$ by summing the probabilities of exactly 3 and exactly 4 successes using the binomial distribution	1
Calculates the correct final probability of $\frac{297}{625}$	1

Q5b

2 marks

In a randomly selected group of four people, what is the probability that exactly two people have the SPGE1 gene, given that at least one of those people has the SPGE1 gene? Express your answer in the form $\frac{a^3}{b^4 - c^4}$ , where $a, b, c \in Z^+$ .

Reveal Answer

$\Pr(X = 2 | X \geq 1) = \frac{\Pr(X = 2) \cap \Pr(X \geq 1)}{\Pr(X \geq 1)}$

$= \frac{\Pr(X = 2)}{1 - \Pr(X = 0)}$

$= \frac{6 \times \left(\frac{3}{5}\right)^2 \times \left(\frac{2}{5}\right)^2}{\left(\frac{5}{5}\right)^4 - \left(\frac{2}{5}\right)^4}$

$= \frac{6^3}{5^4 - 2^4}$

Marking Criteria

Descriptor	Marks
Evaluates $\Pr(X = 2)$ and $\Pr(X \geq 1)$ or correctly substitutes into the conditional probability formula	1
Provides the correct final answer in the required form, $\frac{6^3}{5^4 - 2^4}$	1

Q8

2025

SCSA

Paper 2

8 marks

Q8

A local primary school will be holding a fair to raise money for the school. Mr Gold has proposed a game to be played at the fair. Two red balls and three blue balls are placed in a bag, and the player randomly removes two balls, with the first not being replaced before the second is chosen. A player wins ＄5 for each red ball they remove.

Let the random variable $X$ denote the number of red balls removed from the bag in a single game.

Q8a

2 marks

Complete the probability distribution for $X$ in the table below.

$x$	0	1	2
$\text{P}(X = x)$	$\frac{3}{10}$

Reveal Answer

$x$	0	1	2
$\text{P}(X=x)$	$\frac{3}{10}$	$\frac{2}{5} \times \frac{3}{4} + \frac{3}{5} \times \frac{1}{2} = \frac{3}{5}$	$\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$

Marking Criteria

Descriptor	Marks
correctly calculates first missing probability	1
correctly calculates second missing probability	1

Q8b (i)

1 mark

Calculate the expected value of $X$ .

Reveal Answer

Using CAS

$E(X) = 0.8$

Marking Criteria

Descriptor	Marks
calculates the correct expected value	1

Q8b (ii)

2 marks

Calculate the variance of $X$ .

Reveal Answer

Using CAS

$\sigma(X) = 0.6$

$Var(X) = 0.36$

Marking Criteria

Descriptor	Marks
calculates correct standard deviation	1
calculates the correct variance	1

Q8c

1 mark

Determine the minimum amount that the fair needs to charge for each game so that their expected profit over a large number of games is non-negative.

Reveal Answer

The expected payout for each game is

$＄5 \times \frac{4}{5} = ＄4$

Hence ＄4 is the minimum that the fair needs to charge so that their expected profit is non-negative.

Marking Criteria

Descriptor	Marks
determines the correct minimum cost	1

Q8d

2 marks

Outline how the game would need to be modified so that the number of red balls drawn would be a binomially distributed random variable with a maximum value of 2. Justify your answer.

Reveal Answer

You would need to return the first ball to the bag (and re-mix the balls) prior to the second ball being drawn. This would ensure that the two trials are independent and have the same probability of yielding a red ball.

Marking Criteria

Descriptor	Marks
states that the first ball needs to be returned to the bag before the second ball is drawn	1
justifies this with reference to the independence of the trials and/or each trial having a constant probability of yielding a red ball	1

Q4

2024

QCAA

Paper 2

1 mark

Q4

1 mark

Consider the Bernoulli distribution where the outcomes for rolling a six-sided die are a four and not rolling a four.
Determine the variance of the resulting Bernoulli distribution in this scenario.

A

$0.02\dot{7}$

B

$0.13\dot{8}$

C

$0.1\dot{6}$

D

$0.8\dot{3}$

Reveal Answer

A

$0.02\dot{7}$

This value represents $p^2 = (\frac{1}{6})^2 = \frac{1}{36} \approx 0.02\dot{7}$ . The correct formula for the variance of a Bernoulli distribution is $p(1-p)$ .

B

$0.13\dot{8}$

Correct Answer

The probability of rolling a four is $p = \frac{1}{6}$ . The variance is calculated as $p(1-p) = \frac{1}{6} \cdot \frac{5}{6} = \frac{5}{36}$ , which equals $0.13\dot{8}$ .

C

$0.1\dot{6}$

This is the probability of success, $p = \frac{1}{6} \approx 0.1\dot{6}$ , rather than the variance of the distribution.

D

$0.8\dot{3}$

This is the probability of failure (not rolling a four), $1-p = \frac{5}{6} \approx 0.8\dot{3}$ , rather than the variance.

Q8

2020

VCAA

Paper 2

1 mark

Q8

1 mark

Items are packed in boxes of 25 and the mean number of defective items per box is 1.4

Assuming that the probability of an item being defective is binomially distributed, the probability that a box contains more than three defective items, correct to three decimal places, is

A

0.037

B

0.048

C

0.056

D

0.114

E

0.162

Reveal Answer

A

0.037

This is the probability of getting exactly 4 defective items, $P(X=4) \approx 0.037$ , rather than the cumulative probability of getting more than 3.

B

0.048

Correct Answer

First, find the probability of a single defective item: $p = \frac{1.4}{25} = 0.056$ . Using the binomial distribution $X \sim \text{Bin}(25, 0.056)$ , the probability of more than 3 defective items is $1 - P(X \le 3) \approx 0.048$ .

C

0.056

This is the probability $p$ of a single item being defective ( $1.4/25 = 0.056$ ), not the probability of finding more than 3 defective items in a box.

D

0.114

This is the probability of getting exactly 3 defective items, $P(X=3) \approx 0.114$ , rather than strictly more than 3.

E

0.162

This is the probability of getting 3 or more defective items, $P(X \ge 3) \approx 0.162$ . The question asks for strictly more than 3, which is $P(X > 3)$ .

Q3

2024

SCSA

Paper 1

6 marks

Q3

The tables below display the partially completed probability distribution and cumulative distribution for a discrete random variable $X$ .

$x$	$1$	$2$	$3$	$4$	$5$
$\text{P}(X = x)$	$0.2$	$0.15$			$0.05$

$x$	$1$	$2$	$3$	$4$	$5$
$\text{P}(X \leq x)$	$0.2$		$0.6$	$0.95$

Q3a

2 marks

Complete the missing probability entries in each of the tables above.

Reveal Answer

$x$	1	2	3	4	5
$P(X=x)$	0.2	0.15	0.25	0.35	0.05

$x$	1	2	3	4	5
$P(X\le x)$	0.2	0.35	0.6	0.95	1

Marking Criteria

Descriptor	Marks
correctly completes two blank entries in the tables	1
correctly completes the remaining two blank entries	1

Q3b

2 marks

Evaluate $\text{P}(2 \leq X \leq 4)$ .

Reveal Answer

$\begin{aligned} P(2\le X\le 4)&=P(X\le 4)-P(X\le 1)\\ &=0.95-0.2\\ &=0.75 \end{aligned}$

Marking Criteria

Descriptor	Marks
writes a correct probability statement in terms of individual/cumulative probabilities	1
calculates correct probability	1

Q3c

2 marks

Evaluate $\text{P}(X = 1 \,|\, X \leq 3)$ .

Reveal Answer

$\begin{aligned} P(X=1\mid X\le 3)&=\dfrac{P(X=1)}{P(X\le 3)}\\ &=\dfrac{0.2}{0.6}\\ &=\dfrac{1}{3} \end{aligned}$

Marking Criteria

Descriptor	Marks
writes a correct probability statement in terms of individual/cumulative probabilities	1
calculates correct probability	1

Q11

2023

QCAA

Paper 1

4 marks

Q11

Two random samples (A and B) were obtained using two different Bernoulli experiments. Each Bernoulli trial in the random samples was recorded as 1 (for success) or 0 (for failure). The results are shown.


A	1	1	1	1	0	1	1	0	1	1
B	0	0	1	1	1	0	1	1	0	0

In sample A, for each trial the mean is 0.8 and the variance is 0.16.

Q11a

2 marks

Use the sample B results to determine the mean and variance for each trial in sample B.

Reveal Answer

Sample B
Mean $= p = \frac{5}{10} = 0.5$

Variance $= p(1-p) = 0.5 \times 0.5$
$= 0.25$

Marking Criteria

Descriptor	Marks
Correctly determines the mean for sample B	1
Determines the variance for sample B	1

Q11b

2 marks

Compare the variability about the means of samples A and B.

Reveal Answer

Sample B has a larger variance than sample A.
Because sample B has the larger variance, it has more variability about the mean compared to sample A.

Marking Criteria

Descriptor	Marks
Identifies that sample B has the larger variance	1
Provides reasoning that explains that larger variance indicates larger variability	1

Q2

2023

QCAA

Paper 2

1 mark

Q2

1 mark

The probability of hitting a bullseye on a standard dartboard is 1 in 1250. What is the probability of hitting a bullseye exactly once in 10 attempts?

A

$\binom{9}{1}\left(\frac{1}{1250}\right)^1 \times \left(\frac{1249}{1250}\right)^9$

B

$\binom{9}{1}\left(\frac{1}{1250}\right)^9 \times \left(\frac{1249}{1250}\right)^1$

C

$\binom{10}{1}\left(\frac{1}{1250}\right)^1 \times \left(\frac{1249}{1250}\right)^9$

D

$\binom{10}{1}\left(\frac{1}{1250}\right)^9 \times \left(\frac{1249}{1250}\right)^1$

Reveal Answer

A

$\binom{9}{1}\left(\frac{1}{1250}\right)^1 \times \left(\frac{1249}{1250}\right)^9$

This option incorrectly uses $n=9$ for the number of trials in the binomial coefficient $\binom{n}{k}$ . Since there are 10 attempts, the coefficient should be $\binom{10}{1}$ .

B

$\binom{9}{1}\left(\frac{1}{1250}\right)^9 \times \left(\frac{1249}{1250}\right)^1$

This option uses the wrong number of trials ( $n=9$ ) and swaps the exponents, calculating the probability of 9 successes and 1 failure instead of 1 success and 9 failures.

C

$\binom{10}{1}\left(\frac{1}{1250}\right)^1 \times \left(\frac{1249}{1250}\right)^9$

Correct Answer

This correctly applies the binomial probability formula $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$ with $n=10$ trials, $k=1$ success, and probability of success $p=\frac{1}{1250}$ .

D

$\binom{10}{1}\left(\frac{1}{1250}\right)^9 \times \left(\frac{1249}{1250}\right)^1$

This option swaps the exponents for success and failure. It calculates the probability of hitting the bullseye 9 times ( $k=9$ ) and missing once, rather than hitting it exactly once.

Q13

2023

QCAA

Paper 1

5 marks

Q13

At a certain airport, the departure of one in five international flights is delayed every day. The status of any flight is independent of other flights.
One international flight is selected at random each day for three days. Each selection is recorded as either 'delayed' or 'not delayed'.

Q13a

2 marks

State two conditions that make this context suitable for modelling using a binomial random variable.

Reveal Answer

One condition is that there are only two outcomes for each selection: 'delayed' (success) or 'not delayed' (failure), i.e. Bernoulli trials.
Another condition that makes this context suitable is that the probabilities of each outcome do not change in each trial, i.e. $p = \frac{1}{5}$ and $q = \frac{4}{5}$

Marking Criteria

Descriptor	Marks
Correctly states one condition for binomial probability	1
Correctly states a second condition for binomial probability	1

Q13b

3 marks

Calculate the probability that at least two of the selected flights were delayed.

Reveal Answer

$n=3, p=\frac{1}{5}, 1-p=\frac{4}{5}$
$P(\text{at least } 2)$
$= P(X=2 \text{ or } X=3))$
$= C_2^3 p^2 (1-p)^1 + C_3^3 p^3 (1-p)^0$
$= 3 \left(\frac{1}{5}\right)^2 \times \left(\frac{4}{5}\right) + \left(\frac{1}{5}\right)^3$
$= 3 \times \frac{1}{25} \times \frac{4}{5} + \frac{1}{5^3}$
$= \frac{12}{125} + \frac{1}{125}$
$= \frac{13}{125}$

Marking Criteria

Descriptor	Marks
Correctly determines the number of trials and the probability of a flight being delayed	1
Determines a suitable method	1
Calculates the probability	1

Q4

2022

VCAA

Paper 1

5 marks

Q4

A card is drawn from a deck of red and blue cards. After verifying the colour, the card is replaced in the deck. This is performed four times.

Each card has a probability of $\frac{1}{2}$ of being red and a probability of $\frac{1}{2}$ of being blue.

The colour of any drawn card is independent of the colour of any other drawn card.

Let $X$ be a random variable describing the number of blue cards drawn from the deck, in any order.

Q4a

2 marks

Complete the table below by giving the probability of each outcome.

$x$	0	1	2	3	4
$\text{Pr}(X = x)$	$\frac{1}{16}$		$\frac{6}{16}$

Reveal Answer

$x$	0	1	2	3	4
$\text{Pr}(X = x)$	$\frac{1}{16}$	$\frac{4}{16}$	$\frac{6}{16}$	$\frac{4}{16}$	$\frac{1}{16}$

Marking Criteria

Descriptor	Marks
Correctly completes the table with $\frac{4}{16}$ , $\frac{4}{16}$ and $\frac{1}{16}$ .	2
Correctly identifies at least one missing probability.	1
No correct probabilities identified.	0

Q4b

1 mark

Given that the first card drawn is blue, find the probability that exactly two of the next three cards drawn will be red.

Reveal Answer

$\frac{3}{8}$

Marking Criteria

Descriptor	Marks
Correct answer of $\frac{3}{8}$ .	1

Q4c

2 marks

The deck is changed so that the probability of a card being red is $\frac{2}{3}$ and the probability of a card being blue is $\frac{1}{3}$ .

Given that the first card drawn is blue, find the probability that exactly two of the next three cards drawn will be red.

Reveal Answer

$\text{Pr}(2Red | First Blue)$
$= \text{Pr}(RRB) + \text{Pr}(RBR) + \text{Pr}(BRR)$
$= 3\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^1$
$= \frac{4}{9}$

Marking Criteria

Descriptor	Marks
Correct working, such as $3\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^1$ or an equivalent correct conditional probability setup.	1
Correct answer of $\frac{4}{9}$ .	1

Q8

2023

VCAA

Paper 2

1 mark

Q8

1 mark

A box contains $n$ green balls and $m$ red balls. A ball is selected at random, and its colour is noted. The ball is then replaced in the box.

In 8 such selections, where $n \neq m$ , what is the probability that a green ball is selected at least once?

A

$8\left(\frac{n}{n+m}\right)\left(\frac{m}{n+m}\right)^7$

B

$1 - \left(\frac{n}{n+m}\right)^8$

C

$1 - \left(\frac{m}{n+m}\right)^8$

D

$1 - \left(\frac{n}{n+m}\right)\left(\frac{m}{n+m}\right)^7$

E

$1 - 8\left(\frac{n}{n+m}\right)\left(\frac{m}{n+m}\right)^7$

Reveal Answer

A

$8\left(\frac{n}{n+m}\right)\left(\frac{m}{n+m}\right)^7$

This formula calculates the probability of selecting exactly one green ball in 8 draws, rather than at least one green ball.

B

$1 - \left(\frac{n}{n+m}\right)^8$

This calculates the probability of selecting at least one red ball, which is found by subtracting the probability of selecting all green balls from 1.

C

$1 - \left(\frac{m}{n+m}\right)^8$

Correct Answer

The probability of selecting at least one green ball is equal to 1 minus the probability of selecting no green balls (all red balls). Since the probability of selecting a red ball is $\frac{m}{n+m}$ , the probability of 8 red balls is $\left(\frac{m}{n+m}\right)^8$ .

D

$1 - \left(\frac{n}{n+m}\right)\left(\frac{m}{n+m}\right)^7$

This subtracts the probability of getting one specific sequence of 1 green and 7 red balls from 1, which does not represent the probability of getting at least one green ball.

E

$1 - 8\left(\frac{n}{n+m}\right)\left(\frac{m}{n+m}\right)^7$

This calculates the probability of NOT selecting exactly one green ball, which is 1 minus the probability of getting exactly one green ball.

QCAA Mathematical Methods Discrete random variables

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