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Probability, MET2 2023 VCAA 8 MC

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?

  1. \(8\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7\)
  2. \(1-\Bigg(\dfrac{n}{n+m}\Bigg)^8\)
  3. \(1-\Bigg(\dfrac{m}{n+m}\Bigg)^8\)
  4. \(1-\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7\)
  5. \(1-8\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Let}\ \ X=\ \text{choosing a green ball}\)

\(\text{Pr}(X\geq 1)\) \(=1-\text{Pr}(X=0)\)
  \(=1-\Bigg(\dfrac{m}{n+m}\Bigg)^8\)

 
\(\Rightarrow C\)


♦ Mean mark 49%.

Probability, MET2-NHT 2019 VCAA 9 MC

At the start of a particular week, Kim has three red apples and two green apples. She eats one apple everyday. On Monday, Tuesday and Wednesday of that week, she randomly selects an apple to eat. In this three-day period, the probability that Kim does not eat an apple of the same colour on any two consecutive days is

  1.  `(1)/(10)`
  2.  `(1)/(5)`
  3.  `(3)/(10)`
  4.  `(2)/(5)`
  5.  `(6)/(25)`
Show Answers Only

`C`

Show Worked Solution

`text{Pr (alternate colours)}`

`= text(Pr)(RGR) + text(Pr)(GRG)`

`= (3)/(5) ·(2)/(4) ·(2)/(3) + (2)/(5) ·(3)/(4) ·(1)/(3)`

`= (12)/(60) + (6)/(60)`

`= (3)/(10)`
 

`=> \ C`

Probability, MET1-NHT 2019 VCAA 6a

Jacinta tosses a coin five times.

Assuming that the coin is fair and given that Jacinta observes a head on the first two tosses, find the probability that she observes a total of either four or five heads.  (2 marks)

Show Answers Only

`1/2`

Show Worked Solution

`text(After 2 tosses, 2 heads.)`

`text(Pr)text{(4 or 5}\ H)` `= HHT + HTH + THH + HHH`
  `= (1/2)^3 xx 4`
  `= 1/2`

Probability, MET2 2018 VCAA 13 MC

In a particular scoring game, there are two boxes of marbles and a player must randomly select one marble from each box. The first box contains four white marbles and two red marbles. The second box contains two white marbles and three red marbles. Each white marble scores −2 points and each red marble scores +3 points. The points obtained from the two marbles randomly selected by a player are added together to obtain a final score.

What is the probability that the final score will equal +1?

  1. `2/3`
  2. `1/5`
  3. `2/5`
  4. `2/15`
  5. `8/15`
Show Answers Only

`E`

Show Worked Solution

`text(Final score requires a red and white.)`

`:. P(+1)` `= P(WR) + P(RW)`
  `= 4/6 xx 3/5 + 2/6 xx 2/5`
  `= 8/15`

 
`=>   E`

Probability, MET2 2017 VCAA 3 MC

A box contains five red marbles and three yellow marbles. Two marbles are drawn at random from the box without replacement.

The probability that the marbles are of different colours is

  1. `5/8`
  2. `3/5`
  3. `15/28`
  4. `15/56`
  5. `30/28`
Show Answers Only

`C`

Show Worked Solution

 

`text(Pr)\ (RY) + text(Pr)\ (YR)`

`= 5/8 xx 3/7 +3/8 xx 5/7`

`= 15/28`

`=> C`

Probability, MET2 2011 VCAA 2*

In a chocolate factory the material for making each chocolate is sent to one of two machines, machine A or machine B.

The time, `X` seconds, taken to produce a chocolate by machine A, is normally distributed with mean 3 and standard deviation 0.8.

The time, `Y` seconds, taken to produce a chocolate by machine B, has the following probability density function
 

`f(y) = {{:(0,y < 0),(y/16,0 <= y <= 4),(0.25e^(−0.5(y-4)),y > 4):}`
 

  1. Find correct to four decimal places

    1. `text(Pr)(3 <= X <= 5)`   (1 mark)

    2. `text(Pr)(3 <= Y <= 5)`   (3 marks)

  2. Find the mean of `Y`, correct to three decimal places.   (3 marks)

  3. It can be shown that  `text(Pr)(Y <= 3) = 9/32`. A random sample of 10 chocolates produced by machine B is chosen. Find the probability, correct to four decimal places, that exactly 4 of these 10 chocolate took 3 or less seconds to produce.   (2 marks)

All of the chocolates produced by machine A and machine B are stored in a large bin. There is an equal number of chocolates from each machine in the bin.

It is found that if a chocolate, produced by either machine, takes longer than 3 seconds to produce then it can easily be identified by its darker colour.

  1. A chocolate is selected at random from the bin. It is found to have taken longer than 3 seconds to produce.
  2. Find, correct to four decimal places, the probability that it was produced by machine A.   (3 marks)

Show Answers Only

a.i.  `0.4938`

a.ii. `0.4155`

b.    `4.333`

c.    `0.1812`

d.    `0.4103`

Show Worked Solution

a.i.   `X ∼\ N(3,0.8^2)`

`text(Pr)(3 <= X <= 5) = 0.4938\ \ text{(4 d.p.)}`

 

a.ii.    `text(Pr)(3 <= Y <= 5)` `= text(Pr)(3 <= Y <= 4) + (4 < Y <= 5)`
    `= int_3^4 (y/16)dy + int_4^5(1/4 e^(−1/2(y-4)))dy`
    `= 0.4155\ \ text{(4 d.p.)}`

 

b.    `text(E)(Y)` `= int_0^4 y(y/16)dy + int_4^∞ 0.25ye^(−1/2(y-4))dy`
    `= 4.333\ \ text{(3 d.p.)}`

 

c.   `text(Solution 1)`

`text(Let)\ \ W = #\ text(chocolates from)\ B\ text(that take less)`

`text(than 3 seconds)`

`W ∼\ text(Bi)(10, 9/32)`

`text(Using CAS: binomPdf)(10, 9/32,4)`

`text(Pr)(W = 4) = 0.1812\ \ text{(4 d.p.)}`
 

`text(Solution 2)`

`text(Pr)(W = 4)`  `=((10),(4)) (9/32)^4 (23/32)^6` 
  `=0.1812`

♦♦♦ Mean mark part (e) 19%.
MARKER’S COMMENT: Students who used tree diagrams were the most successful.

 

d.   
`text(Pr)(A | L)` `= (text(Pr)(AL))/(text(Pr)(L))`
  `= (0.5 xx 0.5)/(0.5 xx 0.5 + 0.5 xx 23/32)`
  `= 0.4103\ \ text{(4 d.p.)}`

Probability, MET1 2006 VCAA 10

Jo has either tea or coffee at morning break. If she has tea one morning, the probability she has tea the next morning is 0.4. If she has coffee one morning, the probability she has coffee the next morning is 0.3. Suppose she has coffee on a Monday morning. What is the probability that she has tea on the following Wednesday morning?  (3 marks)

Show Answers Only

`0.49`

Show Worked Solution

vcaa-2006-meth-10

`text(Pr)(C\ T\ T) + text(Pr)(C\ C\ T)`

`= 0.7 xx 0.4 + 0.3 xx 0.7`

`= 0.28 + 0.21`

`= 0.49`

Probability, MET1 2008 VCAA 8

Every Friday Jean-Paul goes to see a movie. He always goes to one of two local cinemas – the Dandy or the Cino.

If he goes to the Dandy one Friday, the probability that he goes to the Cino the next Friday is 0.5. If he goes to the Cino one Friday, then the probability that he goes to the Dandy the next Friday is 0.6.

On any given Friday the cinema he goes to depends only on the cinema he went to on the previous Friday.

If he goes to the Cino one Friday, what is the probability that he goes to the Cino on exactly two of the next three Fridays?  (3 marks)

Show Answers Only

`0.336`

Show Worked Solution

`text(Pr)(C CD) + text(Pr)(CDC) + text(Pr)(DC C)`

`= (0.4 xx 0.4 xx 0.6) + (0.4 xx 0.6 xx 0.5) + (0.6 xx 0.5 xx 0.4)`

`= 0.096 + 0.12 + 0.12`

MARKER’S COMMENT:
Most efficient strategy here was listing the three possibilities – not a tree diagram.

`= 0.336`

 

 

Probability, MET2 2014 VCAA 22 MC

John and Rebecca are playing darts. The result of each of their throws is independent of the result of any other throw. The probability that John hits the bullseye with a single throw is `1/4.` The probability that Rebecca hits the bullseye with a single throw is `1/2.` John has four throws and Rebecca has two throws.

The ratio of the probability of Rebecca hitting the bullseye at least once to the probability of John hitting the bullseye at least once is

  1. `1:1`
  2. `32:27`
  3. `64:85`
  4. `2:1`
  5. `192:175`
Show Answers Only

`E`

Show Worked Solution

`X = #\ text(times John hits bullseye.)`

♦ Mean mark 37%.

`Y = #\ text(times Rebecca hits bullseye.)`

`text(Required ratio:)`

`text(Pr)(Y >= 1)` `\ :\ text(Pr)(X >= 1)`
`1-text(Pr)(Y=0)` `\ :\ 1-text(Pr)(X=0)`
`1-(1/2)^2` `\ :\ 1-(3/4)^4`
`3/4` `\ :\ 175/256`
`192/256` `\ :\ 175/256`
`:. 192` `\ :\ 175`

 
`=>   E`