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Date November 2012 Marks available 1 Reference code 12N.3sp.hl.TZ0.4
Level HL only Paper Paper 3 Statistics and probability Time zone TZ0
Command term Find Question number 4 Adapted from N/A

Question

Jenny and her Dad frequently play a board game. Before she can start Jenny has to throw a “six” on an ordinary six-sided dice. Let the random variable X denote the number of times Jenny has to throw the dice in total until she obtains her first “six”.

If the dice is fair, write down the distribution of X , including the value of any parameter(s).

[1]
a.

Write down E(X ) for the distribution in part (a).

[1]
b.

Before Jenny’s Dad can start, he has to throw two “sixes” using a fair, ordinary six-sided dice. Let the random variable Y denote the total number of times Jenny’s Dad has to throw the dice until he obtains his second “six”.

Write down the distribution of Y , including the value of any parameter(s).

[1]
d.

Before Jenny’s Dad can start, he has to throw two “sixes” using a fair, ordinary six-sided dice. Let the random variable Y denote the total number of times Jenny’s Dad has to throw the dice until he obtains his second “six”.

Find the value of y such that \({\text{P}}(Y = y) = \frac{1}{{36}}\).

[1]
e.

Before Jenny’s Dad can start, he has to throw two “sixes” using a fair, ordinary six-sided dice. Let the random variable Y denote the total number of times Jenny’s Dad has to throw the dice until he obtains his second “six”.

Find \({\text{P}}(Y \leqslant 6)\) .

[2]
f.

Markscheme

\(X \sim {\text{Geo}}\left( {\frac{1}{6}} \right){\text{ or NB}}\left( {1,\frac{1}{6}} \right)\)     A1

[1 mark]

a.

\({\text{E}}(X) = 6\)     A1

[1 mark]

b.

Y is \({\text{NB}}\left( {2,\frac{1}{6}} \right)\)     A1

[1 mark]

d.

\({\text{P}}(Y = y) = \frac{1}{{36}}{\text{ gives }}y = 2\)     A1

(as all other probabilities would have a factor of 5 in the numerator)

[1 mark]

e.

\({\text{P}}(Y \leqslant 6) = {\left( {\frac{1}{6}} \right)^2} + 2\left( {\frac{5}{6}} \right){\left( {\frac{1}{6}} \right)^2} + 3{\left( {\frac{5}{6}} \right)^2}{\left( {\frac{1}{6}} \right)^2} + 4{\left( {\frac{5}{6}} \right)^3}{\left( {\frac{1}{6}} \right)^2} + 5{\left( {\frac{5}{6}} \right)^4}{\left( {\frac{1}{6}} \right)^2}\)     (M1)

\( = 0.263\)     A1

[2 marks]

f.

Examiners report

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

a.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

b.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

d.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

e.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

f.

Syllabus sections

Topic 7 - Option: Statistics and probability » 7.1 » Negative binomial distribution.

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