smc-991-30-Independent Events

Probability, 2ADV S1 2023 HSC 31

Four Year 12 students want to organise a graduation party. All four students have the same probability, \(P(F)\), of being available next Friday. All four students have the same probability, \(P(S)\), of being available next Saturday.

It is given that \(P(F)=\dfrac{3}{10}, P(S\mid F)=\dfrac{1}{3}\), and \(P(F\mid S)=\dfrac{1}{8}\).

Kim is one of the four students.

Is Kim's availability next Friday independent from his availability next Saturday? Justify your answer. (1 mark)

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Show that the probability that Kim is available next Saturday is \(\dfrac{4}{5}\). (2 marks)

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What is the probability that at least one of the four students is NOT available next Saturday? (2 marks)

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a. \(P(F) \neq P(F|S)\ \text{which is not the case}\ \ (\dfrac{3}{10} \neq \dfrac{1}{8}) \)

\(\therefore\ \text{Kim’s availability on Friday is not independent of Saturday}\)

b. \(\text{See Worked Solutions} \)

c. \(\dfrac{369}{625} \)

Show Worked Solution

a. \(P(F) \neq P(F|S)\ \text{which is not the case}\ \ (\dfrac{3}{10} \neq \dfrac{1}{8}) \)

\(\therefore\ \text{Kim’s availability on Friday is not independent of Saturday}\)

♦♦♦ Mean mark (a) 18%.

b.	\(P(S\|F) \)	\(= \dfrac{P(S) \cap P(F)}{P(F)} \)
	\(\dfrac{1}{3}\)	\(= \dfrac{P(S) \cap P(F)}{\frac{3}{10}} \)
	\(\dfrac{1}{10}\)	\(=P(S) \cap P(F) \)

♦ Mean mark (b) 44%.

\(P(F\|S)\)	\(= \dfrac{P(F) \cap P(S)}{P(S)} \)
\(\dfrac{1}{8}\)	\(=\dfrac{\frac{1}{10}}{P(S)} \)
\(\dfrac{1}{8} \times P(S) \)	\(=\dfrac{1}{10} \)
\(P(S)\)	\(=\dfrac{4}{5} \)

c.	\(P\text{(at least 1 not available)}\)	\(=1-P\text{(all are available)} \)
		\(=1-(\dfrac{4}{5})^4 \)
		\(=\dfrac{369}{625} \)

♦♦ Mean mark (c) 34%.

Probability, 2ADV S1 2019 MET2 11 MC

`A` and `B` are events from a sample space such that `P(A) = p`, where `p > 0, \ P(B\ text{|}\ A) = m` and `P(B\ text{|}\ A^{′}) = n`.

`A` and `B` are independent events when

`m = n`
`m = 1-p`
`m + n = 1`
`m = p`

Show Answers Only

`A`

Show Worked Solution

`text(S) text(ince)\ A and B\ text(are independent:)`

`P(B\ text{|}\ A) = P(B\ text{|}\ A^{′})`

`:. m = n`

`=> A`

Probability, 2ADV S1 2007 MET1 6

Two events, `A` and `B`, from a given event space, are such that `P(A) = 1/5` and `P(B) = 1/3`.

Calculate `P(A′ ∩ B)` when `P(A ∩ B) = 1/8`. (1 mark)

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Calculate `P(A′ ∩ B)` when `A` and `B` are mutually exclusive events. (1 mark)

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`5/24`
`1/3`

Show Worked Solution

i. `text(Sketch Venn diagram:)`

♦♦ Mean mark 31%.
MARKER’S COMMENTS: Students who drew a Venn diagram or Karnaugh map were the most successful.

`:.P(A′ ∩ B)`	`=P(B) – P(A ∩B)`
	`=1/3 – 1/8`
	`=5/24`

♦♦ Mean mark 23%.

ii.

`text(Mutually exclusive means)\ \ P(A ∩ B)=0,`

`:. P(A′ ∩ B) = 1/3`

Probability, 2ADV S1 2011 MET1 8

Two events, `A` and `B`, are such that `P(A) = 3/5` and `P(B) = 1/4.`

If `A^{′}` denotes the compliment of `A`, calculate `P (A^{′} nn B)` when

`P (A uu B) = 3/4` (2 marks)

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`A` and `B` are mutually exclusive. (1 mark)

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`3/20`
`1/4`

Show Worked Solution

i. `text(Sketch Venn Diagram)`

`P (A uu B)`	`= P (A) + P (B)-P (A nn B)`
`3/4`	`= 3/5 + 1/4-P (A nn B)`
`P (A nn B)`	`= 1/10`

`:.\ P(A^{′} nn B) = 1/4-1/10 = 3/20`

ii.

`P (A∩ B)=0\ \ text{(mutually exclusive)},`

`:.\ P (A^{′} nn B) = P (B) = 1/4`

Probability, 2ADV S1 2015 MET1 8

For events `A` and `B` from a sample space, `P(A | B) = 3/4` and `P(B) = 1/3`.

Calculate `P(A ∩ B)`. (1 mark)

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Calculate `P(A′ ∩ B)`, where `A′` denotes the complement of `A`. (1 mark)

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If events `A` and `B` are independent, calculate `P(A ∪ B)`. (1 mark)

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Show Answers Only

`1/4`
`1/12`
`5/6`

Show Worked Solution

i. `text(Using Conditional Probability:)`

`P(A \| B)`	`= (P(A ∩ B))/(P(B))`
`3/4`	`= (P(A ∩ B))/(1/3)`
`:. P(A ∩ B)`	`= 1/4`

ii.

`P(A′ ∩ B)`	`= P(B) – P(A ∩B)`
	`= 1/3 – 1/4`
	`= 1/12`

iii. `text(If)\ A, B\ text(independent)`

♦♦ Mean mark 28%.
MARKER’S COMMENT: A lack of understanding of independent events was clearly evident.

`P(A ∩ B)`	`= P(A) xx P(B)`
`1/4`	`= P(A) xx 1/3`
`:. P(A)`	`= 3/4`

`P(A ∪ B)`	`= P(A) + P(B) – P(A ∩ B)`
	`= 3/4 + 1/3 – 1/4`
`:. P(A ∪ B)`	`= 5/6`

Probability, 2ADV S1 2010 MET2 21 MC

Events `A` and `B` are mutually exclusive events of a sample space with

`P(A) = p and P (B) = q\ \ text(where)\ \ 0 < p < 1 and 0 < q < 1.`

`P (A^{′} nn B^{′})` is equal to

`(1-p) (1-q)`
`1-pq`
`1-(p + q)`
`1 - (p + q - pq)`

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`C`

Show Worked Solution

♦ Mean mark 43%.

`P (A^{′} nn B^{′})`	`= 1-p-q`
	`= 1-(p + q)`

`=> C`

Probability, 2ADV S1 2013 MET2 10 MC

For events `A` and `B,\ P(A ∩ B) = p,\ P(A^{′}∩ B) = p -1/8` and `P(A ∩ B^{′}) = (3p)/5.`

If `A` and `B` are independent, then the value of `p` is

`0`
`1/4`
`3/8`
`1/2`

Show Answers Only

`C`

Show Worked Solution

`P(A)`	`= P(A ∩ B) + P(A ∩ B^{′})`
	`= p + (3p)/5`
	`= (8p)/5`

`P(B)`	`= P(B ∩ A) + P(B ∩ A^{′})`
	`= p + p-1/8`
	`= 2p – 1/8`

`text(S)text(ince)\ A and B\ text(are independent events,)`

`P(A ∩ B)`	` = P(A) xx P(B)`
`p`	`=(8p)/5 (2p-1/8)`
`5p`	`=16p^2-p`
`16p^2-6p`	`=0`
`2p(8p-3)`	`=0`
`:.p`	`=3/8,\ \ \ p!=0`

`=> C`