smc-829-30-P(E) = 1

Probability, STD2 S2 2024 HSC 31

A coin is biased so that it is twice as likely to show a head than a tail.

What is the probability of obtaining a head with one throw of this coin? (1 mark)
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In two throws of this coin, what is the probability of obtaining at least one head? (2 marks)
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a. \(P(H) = \dfrac{2}{3}\)

b. \(P\text{(at least 1 head)}\ = \dfrac{8}{9}\)

Show Worked Solution

a. \(P(H) = \dfrac{2}{3}\)

b. \(P(T) = 1-P(H)=1-\dfrac{2}{3}=\dfrac{1}{3}\)

	\(P\text{(at least 1 head)}\)	\(=1-P\text{(no heads)}\)
		\(=1-P(TT)\)
		\(=1-\dfrac{1}{3} \times \dfrac{1}{3}\)
		\(=\dfrac{8}{9}\)

♦ Mean mark (a) 51%.
♦♦ Mean mark (b) 30%.

Probability, STD2 S2 SM-Bank 1

A game consists of two tokens being drawn at random from a barrel containing 20 tokens. There are 17 red tokens and 3 black tokens. The player keeps the two tokens drawn.

Complete the probability tree by writing the missing probabilities in the boxes. (2 marks)
What is the probability that a player draws at least one red token. Give your answer in exact form. (2 marks)

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`187/190`

Show Worked Solution

ii. `P(text(at least one red))`

`= 1 – P(BB)`

`= 1 – 3/20 · 2/19`

`= 187/190`

Probability, STD2 S2 2009 HSC 27c

In each of three raffles, 100 tickets are sold and one prize is awarded.

Mary buys two tickets in one raffle. Jane buys one ticket in each of the other two raffles.

Determine who has the better chance of winning at least one prize. Justify your response using probability calculations. (4 marks)

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`P(text(Mary wins) )`	`= 2/100`
	`= 1/50`

`P(text(Jane wins at least 1) )`	`= 1-P (text(loses both) )`
	`= 1-99/100 xx 99/100`
	`= 1-9801/(10\ 000)`
	`= 199/(10\ 000)`

`text{Since}\ \ 1/50 > 199/(10\ 000)`

`=>\ text(Mary has a better chance of winning.)`

Show Worked Solution

`P(text(Mary wins) )`	`= 2/100`
	`= 1/50`

`P(text(Jane wins at least 1) )`	`= 1-P (text(loses both) )`
	`= 1-99/100 xx 99/100`
	`= 1-9801/(10\ 000)`
	`= 199/(10\ 000)`

`text{Since}\ \ 1/50 > 199/(10\ 000)`

`=>\ text(Mary has a better chance of winning.)`

♦♦ Mean mark 31%.
MARKER’S COMMENT: Very few students calculated Jane’s chance of winning correctly. Note the use of “at least” in the question. Finding `1-P`(complement) is the best strategy here.

Probability, STD2 S2 2012 HSC 12 MC

Two unbiased dice, each with faces numbered 1, 2, 3, 4, 5, 6, are rolled.

What is the probability of a 6 appearing on at least one of the dice?

`1/6`
`11/36`
`25/36`
`5/6`

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`B`

Show Worked Solution

`P(text(at least 1 six))`

`= 1-P(text(no six)) xx P(text(no six))`

`=1-5/6 xx 5/6`

`=11/36`

`=> B`

♦♦♦ Mean mark 25%
COMMENT: The term “at least” should flag that calculating the probability of `1-P text{(event not happening)}` is likely to be the most efficient way to solve.