A2 October 2021 Q5

EdexcelCurrent spec18 marksIncludes hypothesis testingGeometric & Negative Binomial Quality of Tests

5. Asha, Davinda and Jerry each have a bag containing a large number of counters, some of which are white and the rest are red.
Each person draws counters from their bag one at a time, notes the colour of the counter and returns it to their bag.

The probability of Asha getting a red counter on any one draw is 0.07

(a) Find the probability that Asha will draw at least 3 white counters before a red counter is drawn. (2)

(b) Find the probability that Asha gets a red counter for the second time on her 9th draw. (2)

The probability of Davinda getting a red counter on any one draw is \(p\).
Davinda draws counters until she gets \(n\) red counters. The random variable \(D\) is the number of counters Davinda draws.

Given that the mean and the standard deviation of \(D\) are 4400 and 660 respectively,

Jerry believes that his bag contains a smaller proportion of red counters than Asha’s bag. To test his belief, Jerry draws counters from his bag until he gets a red counter. Jerry defines the random variable \(J\) to be the number of counters drawn up to and including the first red counter.

(d) Stating your hypotheses clearly and using a 10% level of significance, find the critical region for this test. (5)

Jerry gets a red counter for the first time on his 34th draw.

(e) Giving a reason for your answer, state whether or not there is evidence that Jerry’s bag contains a smaller proportion of red counters than Asha’s bag. (2)

Given that the probability of Jerry getting a red counter on any one draw is 0.011

(f) show that the power of the test is 0.702 to 3 significant figures. (3)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{P}(\text{at least 3 whites}) = (1 - 0.07)^3\) or \(1 - 0.07 - 0.93 \times 0.07 - 0.93^2 \times 0.07\)	M1	1.1b
\(= 0.8043\ldots\) awrt 0.804	A1	1.1b
	(2)

Notes

M1: A correct method to find \(\mathrm{P}(X \geqslant 3)\)

A1: awrt 0.804

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{P}(\text{2nd red on 9}^{\text{th}}\text{ draw}) = \dbinom{8}{1}0.93^7 \times 0.07^2\)	M1	3.3
\(= 0.02358\ldots\) awrt 0.0236	A1	1.1b
	(2)

Notes

M1: For selecting the appropriate model negative binomial or binomial with an extra trial

A1: awrt 0.0236

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{n}{p} = 4400\) and \(\dfrac{n(1 - p)}{p^2} = 660^2\)	M1 A1	3.1b 1.1b
\(1 - p = 99p\) oe	M1	1.1b
\(p = 0.01\)	A1	1.1b
	(4)

Notes

M1: Forming an equation for the mean and variance. At least one correct.

A1: Both equations correct
Allow M1 A1 if both equations correct with the same number subst for \(n\)

M1: Solving the 2 equations leading to \(1 - p = 99p\) oe Allow \(p - p^2 = 99p^2\) ft their 4400 and 660 Allow \(1 - p = 0.15p\)

A1: 0.01

Mark scheme (d)

Scheme	Marks	AO
\(\mathrm{H}_0: p = 0.07 \qquad \mathrm{H}_1: p \lt 0.07\)	B1	2.5
\(J \sim \mathrm{Geo}(0.07)\)	M1	3.3
\(\mathrm{P}(J \geqslant c) \lt 0.1 \Rightarrow (1 - 0.07)^{c - 1} \lt 0.1\)	M1	3.4
\(c - 1 \gt \dfrac{\log 0.1}{\log 0.93}\)	M1	1.1b
\(c \gt 32.72\ldots \quad \therefore \text{CR} \quad J \geqslant 33\)	A1	1.1b
	(5)

Notes

B1: Both hypotheses correct using correct notation allow eg \(p \gt 0.93\) (corrected from the printed mark scheme, whose notes label this mark M1; the scheme awards it as B1)

M1: Realising the need to use \(\mathrm{Geo}(0.07)\) ft their Hypotheses

M1: Using the model to find \(\mathrm{P}(J \geqslant c)\) Condone \((1 - 0.07)^c \lt 0.1\) ft their \(0.07 \neq 0.93\)
ALT \(\mathrm{P}(J \geqslant 32) = 0.1[054\ldots]\) or \(\mathrm{P}(J \geqslant 33) = 0.09[8\ldots]\) Implied by correct CR

M1: For a valid method to solve the inequality or \(\mathrm{P}(J \geqslant 32) = 0.1[054]\) and \(\mathrm{P}(J \geqslant 33) = 0.09[81]\) Implied by correct CR

A1: Correct CR (any letter) A0 if given as a probability statement. Must be integer

Mark scheme (e)

Scheme	Marks	AO
34 is in the Critical region	M1	1.1b
There is evidence to suggest that Jerry’s bag contains a smaller proportion of red counters than Asha’s bag.	A1	2.2b
	(2)

Notes

M1: Comparing 34 with their CR eg \(34 \gt 33\) \(34 \geqslant 33\) or \(\mathrm{P}(J \geqslant 34) = 0.09[12]\)

A1: Fully correct conclusion in context. Allow Jerry’s belief is true. Allow probability for proportion

Mark scheme (f)

Scheme	Marks	AO
Power of test \(= \mathrm{P}(J \geqslant 33 \mid p = 0.011)\)	M1	2.1
\(= (1 - 0.011)^{32}\) oe	M1	1.1b
\(= 0.7019\ldots\)*	A1*	1.1b
	(3)
	(18 marks)

Notes

M1: Realising they need to find P(their CR in (d)) Allow \(1 - \mathrm{P}(J \leqslant 32)\)

M1: For a Correct method. Allow \(1 - 0.2981\ldots\) May be implied by \(0.7019\ldots\) If the CR is incorrect \((1 - 0.011)^{\text{“CR”} - 1}\) or \(1 - \{1 - (1 - 0.011)^{\text{“CR”} - 1}\}\) must be seen

A1*: Only award if both method marks awarded.