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Probability, SMB-015

On a tray there are 12 hard‑centred chocolates `(H)` and 8 soft‑centred chocolates `(S)`. Two chocolates are selected at random. A partially completed probability tree is shown.
 


 

What is the probability of selecting at least one soft-centred chocolate?  (3 marks)

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Show Answers Only

`62/95`

Show Worked Solution

`P(text{at least one}\ S)`

`= 1-P(HH)`

`= 1-(12/20 xx 11/19)`

`= 1-33/95`

`= 62/95`

♦ Mean mark 45%.

Filed Under: Multi-Stage Events Tagged With: num-title-ct-pathb, smc-4238-10-Dependent events, smc-4238-50-Probability trees, smc-4238-70-Complementary events, smc-4238-80-"at least"

Probability, SMB-014

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.

  1.  Complete the probability tree by writing the missing probabilities in the boxes.  (2 marks)
     
     
       

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  2.  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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  1.  
  2. `187/190`
Show Worked Solution
i.   

 

ii.   `P(text(at least one red))`

`= 1-P(BB)`

`= 1-3/20 xx 2/19`

`= 187/190`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-pathb, smc-4238-10-Dependent events, smc-4238-50-Probability trees, smc-4238-70-Complementary events, smc-4238-80-"at least"

Probability, SMB-013

Beib owns three white and five blue T-shirts. He chooses a T-shirt at random for himself and puts it on. He then chooses another T-shirt at random, from the remaining T-shirts, and gives it to his brother.

  1. What is the probability that Beib chooses a blue T-shirt for himself?  (1 mark)

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  2. Complete the tree diagram by writing the correct probability on each branch.  (2 marks)
     
         

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  3. Calculate the probability that both of the T-shirts are the same colour.  (2 marks)

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  1. `5/8`
  2.  
  3. `13/28`
Show Worked Solution

i.    `P(B)= (text(number of blue T-shirts))/(text(total number of T-shirts)) = 5/8`

  
ii.
    

       

 
iii.
 `Ptext((same colour))`

`= P(text(WW)) + P(text(BB))`

`= 3/8 xx 2/7 + 5/8 xx 4/7`

`= 6/56 + 20/56`

`= 13/28`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-10-Dependent events, smc-4238-50-Probability trees

Probability, SMB-012

Two dice are rolled. What is the probability that only one of the dice shows a three?   (2 marks)

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`5/18`

Show Worked Solution

`text{Each dice roll is an independent event.}`

`P(3) = 1/6, \ P\text{(not 3)} = 1-1/6=5/6`

`text{P (Only one 3)}`

`= P text{(3, not 3)} + P text{(not 3, 3)}`

`= 1/6 xx 5/6 + 5/6 xx 1/6`

`= 10/36`

`= 5/18`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events, smc-4238-70-Complementary events

Probability, SMB-011

Tay-Tay has 2 bags of apples.

Bag A contains 4 red apples and 3 green apples. 

Bag B contains 3 red apples and 1 green apple.

Tay-Tay chooses one of the bags randomly and with her eyes closed, takes one of the apples. 

  1. Complete the tree diagram below.   (2 marks)
     

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  2. Determine the probability that Tay-Tay chooses a red apple?   (2 marks) 

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i.    
       

ii.    `37/56`

Show Worked Solution

i.    
       
 

ii.     `Ptext{(red)}` `=1/2 xx 4/7 + 1/2 xx 3/4`
    `=37/56`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-10-Dependent events, smc-4238-50-Probability trees

Probability, SMB-010

Each time she throws a dart, the probability that Gaga hits the dartboard is  `4/7`.

She throws two darts, one after the other.

What is the probability that she misses the dartboard with both darts?   (2 marks)

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`9/49`

Show Worked Solution

`P text{(hits)} = 4/7\ \ =>\ \ P\text{(misses)} = 1-4/7=3/7`

`P text{(misses twice)}` `= 3/7 xx 3/7`
  `= 9/49`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events, smc-4238-70-Complementary events

Probability, SMB-009

Jon spins each pointer 50 times.

 

Each time he added the numbers that the pointers landed on.

His results are shown below.

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textbf{Sum of numbers}\rule[-1ex]{0pt}{0pt} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & \textbf{Total}\\
\hline
\rule{0pt}{2.5ex}\textbf{Number of spins}\rule[-1ex]{0pt}{0pt} & 1 & 2 & 3 & 6 & 8 & 9 & 7 & 5 & 4 & 4 & 1 & \textbf{50} \\
\hline
\end{array}

What percentage of the spins resulted in a sum of 9?   (2 marks)

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`text(10%)`

Show Worked Solution
`text(Percentage)` `= text(number of 9’s)/text(number of spins) xx 100`
  `=5/50 xx 100`
  `=10 text(%)`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events, smc-4238-60-Arrays

Probability, SMB-008

Two fair 20 cent coins are tossed at the same time.

What is the probability that both coins will show heads?   (2 marks)

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`1/4`

Show Worked Solution

`text(Possible outcomes are:)`

`text(HH, HT, TH, TT.)`

`:. P text{(HH)} = 1/4`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events

Probability, SMB-007 MC

Brandi spins the arrow on two identical spinners.

The arrow on each spinner is equally likely to land on 1, 2 or 3.
 

If Brandi adds up the two results, which total is she least likely to get?

  1. `2`
  2. `3`
  3. `4`
  4. `5`
Show Answers Only

`A`

Show Worked Solution

`text{2 can only result from (1, 1).}`

`text(All other totals can have more than 1 combination producing them.)`

`=>A`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events

Probability, SMB-006

Bromley rolls two standard dice at the same time and adds up the total.

 An incomplete table of the possible outcomes is below.

\begin{array} {|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textbf{Face}  \rule[-1ex]{0pt}{0pt} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6}\\
\hline
\rule{0pt}{2.5ex}\textbf{1}\rule[-1ex]{0pt}{0pt} & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
\rule{0pt}{2.5ex}\textbf{2}\rule[-1ex]{0pt}{0pt} & 3 & 4 & 5 & & & 8 \\
\hline
\rule{0pt}{2.5ex}\textbf{3}\rule[-1ex]{0pt}{0pt} & 4 & 5 & 6 &  &  & 9 \\
\hline
\rule{0pt}{2.5ex}\textbf{4}\rule[-1ex]{0pt}{0pt} & 5 &  & 7 & 8 & 9 & 10 \\
\hline
\rule{0pt}{2.5ex}\textbf{5}\rule[-1ex]{0pt}{0pt} & 6 &  & 8 &  & 10 & 11 \\
\hline
\rule{0pt}{2.5ex}\textbf{6}\rule[-1ex]{0pt}{0pt} & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\end{array}

  1. Complete the table.   (2 marks)

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  2. What is the most likely total that Bromley will roll?   (1 mark)

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i.

\begin{array} {|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textbf{Face}  \rule[-1ex]{0pt}{0pt} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6}\\
\hline
\rule{0pt}{2.5ex}\textbf{1}\rule[-1ex]{0pt}{0pt} & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
\rule{0pt}{2.5ex}\textbf{2}\rule[-1ex]{0pt}{0pt} & 3 & 4 & 5 & \colorbox{lightblue}{6} & \colorbox{lightblue}{7} & 8 \\
\hline
\rule{0pt}{2.5ex}\textbf{3}\rule[-1ex]{0pt}{0pt} & 4 & 5 & 6 & \colorbox{lightblue}{7} & \colorbox{lightblue}{8} & 9 \\
\hline
\rule{0pt}{2.5ex}\textbf{4}\rule[-1ex]{0pt}{0pt} & 5 & \colorbox{lightblue}{6}  & 7 & 8 & 9 & 10 \\
\hline
\rule{0pt}{2.5ex}\textbf{5}\rule[-1ex]{0pt}{0pt} & 6 & \colorbox{lightblue}{7} & 8 & \colorbox{lightblue}{9} & 10 & 11 \\
\hline
\rule{0pt}{2.5ex}\textbf{6}\rule[-1ex]{0pt}{0pt} & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\end{array}

ii.    `7`

Show Worked Solution

i.

\begin{array} {|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textbf{Face}  \rule[-1ex]{0pt}{0pt} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6}\\
\hline
\rule{0pt}{2.5ex}\textbf{1}\rule[-1ex]{0pt}{0pt} & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
\rule{0pt}{2.5ex}\textbf{2}\rule[-1ex]{0pt}{0pt} & 3 & 4 & 5 & \colorbox{lightblue}{6} & \colorbox{lightblue}{7} & 8 \\
\hline
\rule{0pt}{2.5ex}\textbf{3}\rule[-1ex]{0pt}{0pt} & 4 & 5 & 6 & \colorbox{lightblue}{7} & \colorbox{lightblue}{8} & 9 \\
\hline
\rule{0pt}{2.5ex}\textbf{4}\rule[-1ex]{0pt}{0pt} & 5 & \colorbox{lightblue}{6}  & 7 & 8 & 9 & 10 \\
\hline
\rule{0pt}{2.5ex}\textbf{5}\rule[-1ex]{0pt}{0pt} & 6 & \colorbox{lightblue}{7} & 8 & \colorbox{lightblue}{9} & 10 & 11 \\
\hline
\rule{0pt}{2.5ex}\textbf{6}\rule[-1ex]{0pt}{0pt} & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\end{array}

 
ii.
    `text{The most frequent total = 7 (6 times)}`

`:.\ text{7 is the most likely to be rolled.}`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-60-Arrays

Probability, SMB-005 MC

Two standard dice are rolled at the same time and the two numbers are added up.
 

nap-159-nc03 

Which total is most likely?

  1. `2`
  2. `3`
  3. `4`
  4. `6`
Show Answers Only

`D`

Show Worked Solution

`text(A total of 6 is most likely.)`

`text{Note that a 6 can occur in the following ways:}`

`(5,1), (1,5), (4,2), (2,4) and (3,3)`

`text(No other option given has as many combinations.)`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events

Probability, SMB-004 MC

Peter has a bag of marbles. 75% of his marbles are blue.

Peter takes a green marble from his bag and loses it in a game.

If he takes another marble from the bag without looking, what are the chances it is blue?

  1. less than 75%
  2. equal to 75%
  3. greater than 75%
Show Answers Only

`C`

Show Worked Solution

`text{Taking marbles two times without replacement → dependent events.}`

`text(When taking the second marble, there will be greater than 75% chance)`

`text(of choosing blue because there are the same amount of  blue marbles)`

`text(to be chosen but 1 less marble of another colour.)`

`=>C`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-10-Dependent events

Probability, SMB-003 MC

Arun flips an unbiased coin 200 times.

Which result is most likely?

  1. `20\ text(tails)`
  2. `98\ text(tails)`
  3. `108\ text(tails)`
  4. `196\ text(tails)`
Show Answers Only

`B`

Show Worked Solution

`text{Each toss is an independent event with 50% chance for both heads and tails}.`

`text{The expected result after 200 tosses is 100 tails, 100 heads.}`

`:.\ text{The most likely result = 98 tails (closest to 100)}`

`=> B`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events

Probability, SMB-002

A representative soccer team is chosen from 30 players who play for two clubs, Portland and Lithgow.

10 players from Portland and 20 players from Lithgow are playing in the trials, and 7 players from Portland and 8 from Lithgow are selected in the representative team.
 

 
One player at the trial is randomly selected.

What is the probability that the player is from Lithgow and is selected in the representative team?

Give your answer to two decimal places.   (2 marks)

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Show Answers Only

`0.27`

Show Worked Solution

`Ptext{(player is from Lithgow and selected)}`

`= P(A and B)`

`= P(A) xx P(B)\ \ \ \text{(independent events)}`

`= 20/30 xx 8/20`

`= 8/30`

`= 0.27\ \text{(2 d.p.)}`

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events, smc-4238-50-Probability trees

Probability, SMB-001

A coin is tossed 3 times. There are 8 possible outcomes.

What is the probability of getting 2 heads and 1 tail in any order?   (2 marks)

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Show Answers Only

`3/8`

Show Worked Solution

`text(Strategy 1)`

`text{The table (array) below lists the possible outcomes:}`
 

  1st toss H H H H T T T T
  2nd toss H H T T H H T T
  3rd toss H T H T H T H T
       ✓  ✓    ✓      

`text(From table,)\ Ptext{(2H, 1T)} = 3/8`
  

`text(Strategy 2)`

`text(Using a probability tree:)`
 

Filed Under: Multi-Stage Events Tagged With: num-title-ct-corea, smc-4238-20-Independent events, smc-4238-50-Probability trees, smc-4238-60-Arrays

Probability, STD2 S2 2019 HSC 25

A bowl of fruit contains 17 apples of which 9 are red and 8 are green.

Dennis takes one apple at random and eats it. Margaret also takes an apple at random and eats it.

By drawing a probability tree diagram, or otherwise, find the probability that Dennis and Margaret eat apples of the same colour.  (3 marks)

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Show Answers Only

`8/17`

Show Worked Solution

`P(text(same colour))` `= P(R R) + P(GG)`
  `= 9/17 xx 8/16 + 8/17 xx 7/16`
  `= 72/272 + 56/272`
  `= 8/17`
♦♦ Mean mark 35%.

Filed Under: Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 5, num-title-ct-corea, num-title-qs-hsc, smc-1135-10-Probability Trees, smc-1135-15-Draw Probability Tree, smc-4238-10-Dependent events, smc-4238-50-Probability trees, smc-829-10-Probability Trees, smc-829-15-Draw Probability Tree

Probability, STD2 S2 2015 HSC 16 MC

The probability of winning a game is `7/10`.

Which expression represents the probability of winning two consecutive games?

  1. `7/10 xx 6/9`
  2. `7/10 xx 6/10`
  3. `7/10 xx 7/9`
  4. `7/10 xx 7/10`
Show Answers Only

`D`

Show Worked Solution

`text{Since the two events are independent:}`

`P text{(W)}` `= 7/10`
`P text{(WW)}` `= 7/10 xx 7/10`

 
`=>D`

Filed Under: Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 4, num-title-ct-corea, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-4238-20-Independent events, smc-829-20-Other Multi-Stage Events

Probability, STD2 S2 2005 HSC 23a

There are 100 tickets sold in a raffle. Justine sold all 100 tickets to five of her friends. The number of tickets she sold to each friend is shown in the table.
 

  1. Justine claims that each of her friends is equally likely to win first prize.

     

    Give a reason why Justine’s statement is NOT correct.   (1 mark)

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  2. What is the probability that first prize is NOT won by Khalid or Herman?   (2 marks)

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Show Answers Only
  1. `text(The claim is incorrect because each of her friends)`
    `text(bought a different number of tickets and therefore)`
    `text(their chances of winning are different.)`
  2. `69/100`
Show Worked Solution

i.    `text(The claim is incorrect because each of her friends bought)`

`text(a different number of tickets and therefore their chances of)`

`text(winning are different.)`

 

ii.  `text(Number of tickets not sold to K or H)`

`= 45 + 10 + 14`

`= 69`
 

`:.\ text(Probability 1st prize NOT won by K or H)`

`= 69/100`

Filed Under: Combinations and Single Stage Events (Std 2), Fundamental understanding, Fundamental Understanding (Std 1), Fundamental Understanding (Std 2), Multi-Stage Events, Single and Multi-Stage Events (Std 1), Single stage events Tagged With: Band 3, Band 4, num-title-ct-corea, num-title-qs-hsc, smc-1135-05-Simple Probability, smc-4238-70-Complementary events, smc-828-10-Simple Probability

Probability, STD2 S2 2007 HSC 25c

In a stack of 10 DVDs, there are 5 rated PG, 3 rated G and 2 rated M.

  1. A DVD is selected at random. What is the probability that it is rated M?   (1 mark)

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Grant chooses two DVDs at random from the stack. Copy or trace the tree diagram into your writing booklet.
 

  1. Complete the tree diagram by writing the correct probability on each branch.   (2 marks)

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  2. Calculate the probability that Grant chooses two DVDs with the same rating.   (2 marks)

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Show Answers Only
  1. `1/5`
  2.  
  3. `14/45`
Show Worked Solution

i.    `text(5 PG, 3 G, 2 M)`

`P text{(M)} = 2/10 = 1/5` 

 

ii.   

 

iii.  `P text{(same rating)}`

`= P text{(PG, PG)} + P text{(G, G)} + P text{(M, M)}`

`= (1/2 xx 4/9) + (3/10 xx 2/9) + (1/5 xx 1/9)`

`= 2/9 + 1/15 + 1/45`

`= 14/45`

Filed Under: Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 3, Band 5, num-title-ct-corea, num-title-qs-hsc, smc-1135-10-Probability Trees, smc-4238-10-Dependent events, smc-4238-50-Probability trees, smc-829-10-Probability Trees

Probability, STD2 S2 2014 HSC 16 MC

In Mathsville, there are on average eight rainy days in October.

Which expression could be used to find a value for the probability that it will rain on two consecutive days in October in Mathsville?

  1. `8/31 xx 7/30`
  2. `8/31 xx 7/31`
  3. `8/31 xx 8/30`
  4. `8/31 xx 8/31`
Show Answers Only

`D`

Show Worked Solution

`P text{(rains)} = 8/31\ \ \text{(independent event for each day)}`

`text{Since each day has same probability:}`

`P(R_1 R_2) = 8/31 xx 8/31`

`=>  D`

♦♦♦ Mean mark 16%.
Lowest mark of any MC question in 2014!

Filed Under: Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2) Tagged With: Band 6, num-title-ct-corea, num-title-qs-hsc, smc-4238-20-Independent events, smc-829-20-Other Multi-Stage Events

Probability, STD2 S2 2011 HSC 15 MC

An unbiased coin is tossed 10 times.

A tail is obtained on each of the first 9 tosses.

What is the probability that a tail is obtained on the 10th toss?

  1. `1/2^10`
  2. `1/2`
  3. `1/10`
  4. `9/10`
Show Answers Only

`B`

Show Worked Solution

`text(Each toss is an independent event and has an even chance)`

`text(of being a head or tail.)`

`=> B`

Filed Under: Fundamental understanding, Fundamental Understanding (Std 1), Fundamental Understanding (Std 2), Multi-Stage Events Tagged With: Band 3, num-title-ct-corea, num-title-qs-hsc, smc-4238-20-Independent events

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)  

Show Answers Only
`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.

Filed Under: Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 5, num-title-ct-pathb, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-1135-30-P(E) = 1 - P(not E), smc-4238-70-Complementary events, smc-4238-80-"at least", smc-829-20-Other Multi-Stage Events, smc-829-30-P(E) = 1 - P(not E)

Probability, STD2 S2 2013 HSC 26c

The probability that Michael will score more than 100 points in a game of bowling is `31/40`. 

  1. A commentator states that the probability that Michael will score less than 100 points in a game of bowling is  `9/40`.

     

    Is the commentator correct? Give a reason for your answer.   (1 mark)

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  2. Michael plays two games of bowling. What is the probability that he scores more than 100 points in the first game and then again in the second game?   (1 mark)

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Show Answers Only
  1. `text{Incorrect. Less than “or equal to 100” is correct.}`
  2. `961/1600`
Show Worked Solution
♦♦♦ Mean mark 11%

i.   `text(The commentator is incorrect. The correct)`

`text(statement is)\ Ptext{(score} <=100 text{)} =9/40`

`text{(i.e. less than “or equal to 100” is the correct statement)}`

 

♦ Mean mark 34%
ii. `\ \ \ P(text{score >100 in both})` `= 31/40 xx 31/40` 
    `= 961/1600`

Filed Under: Fundamental understanding, Fundamental Understanding (Std 1), Fundamental Understanding (Std 2), Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 5, Band 6, num-title-ct-corea, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-4238-20-Independent events, smc-4238-70-Complementary events, smc-829-20-Other Multi-Stage Events

Probability, STD2 S2 2010 HSC 20 MC

Lou and Ali are on a fitness program for one month. The probability that Lou will finish the program successfully is 0.7 while the probability that Ali will finish successfully is 0.6. The probability tree shows this information

 

What is the probability that only one of them will be successful ?

  1. `0.18`
  2. `0.28`
  3. `0.42`
  4. `0.46`
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ Ptext{(Lou successful)}=P(L) = 0.7, \ P(\text{not}\ L) = 0.3`

`text(Let)\ \ Ptext{(Ali successful)}=P(A) = 0.6, \ P(\text{not}\ A) = 0.4`

`P text{(only 1 successful)}` `=P(L)xxP(text(not)\ A)+P(text(not)\ L)xxP(A)`
  `=(0.7xx0.4)+(0.3xx0.6)`
  `=0.28+0.18`
  `=0.46`

 
`=>  D`

♦ Mean mark 48%.

Filed Under: Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 5, num-title-ct-corea, num-title-qs-hsc, smc-1135-10-Probability Trees, smc-4238-20-Independent events, smc-4238-50-Probability trees, smc-829-10-Probability Trees

Probability, STD2 S2 2012 HSC 27e

A box contains 33 scarves made from two different fabrics. There are 14 scarves made from silk (S) and 19 made from wool (W).
Two girls each select, at random, a scarf to wear from the box.

  1. Complete the probability tree diagram below.   (2 marks) 
      
       

    --- 0 WORK AREA LINES (style=lined) ---

  2. Calculate the probability that the two scarves selected are made from silk.    (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

  3. Calculate the probability that the two scarves selected are made from different fabrics.   (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

Show Answers Only
  1.  
       
     
  2. `P\ text{(2 silk)}= 91/528`
  3. `P\ text{(different fabrics)}= 133/264`
Show Worked Solution

i. 

♦ Mean mark (i) 43%.
ii.  `P\ text{(2 silk)}` `= P(S_1) xx P(S_2)`
  `= 14/33 xx 13/32`
  `= 91/528`

 

iii.  `P\ text{(different)}` `= P (S_1,W_2) + P(W_1,S_2)`
  `= (14/33 xx 19/32) + (19/33 xx 14/32)`
  `= 532/1056`
  `= 133/264`
♦ Mean mark (iii) 41%.
MARKER’S COMMENT: In better responses, students multiplied along the branches and then added these two results together

Filed Under: Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 4, Band 5, num-title-ct-corea, num-title-qs-hsc, smc-1135-10-Probability Trees, smc-4238-10-Dependent events, smc-4238-50-Probability trees, smc-829-10-Probability Trees

Probability, STD2 S2 2013 HSC 18 MC

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

What is the probability of obtaining a sum of 6?

  1. `1/6`
  2. `1/12`
  3. `5/12`
  4. `5/36`
Show Answers Only

`D`

Show Worked Solution

`text(Total outcomes)=6xx6=36`

`text{Outcomes that sum to 6}=text{(1,5) (5,1) (2,4) (4,2) (3,3)} =5`

`:.\ P\text{(sum of 6)} =5/36`

`=>\ D`

♦♦ Mean mark 35%.

Filed Under: Multi-stage Events, Multi-Stage Events, Multi-Stage Events (Std 2), Single and Multi-Stage Events (Std 1) Tagged With: Band 5, num-title-ct-corea, num-title-qs-hsc, smc-1135-20-Other Multi-Stage Events, smc-4238-20-Independent events, smc-829-20-Other Multi-Stage Events

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