A2 June 2022 Q4

EdexcelCurrent spec13 marksGeometric & Negative Binomial

4. In a game a spinner is spun repeatedly. When the spinner is spun, the probability of it landing on blue is 0.11

(a) Find the probability that the spinner lands on blue

(i) for the first time on the 6th spin, (2)

(ii) for the first time before the 6th spin, (2)

(iii) exactly 4 times during the first 6 spins, (2)

(iv) for the 4th time on or before the 6th spin. (4)

Zac and Izana play the game. They take turns to spin the spinner. The winner is the first one to have the spinner land on blue. Izana spins the spinner first.

(b) Show that the probability of Zac winning is 0.471 to 3 significant figures. (3)

Mark scheme (a)

Scheme	Marks	AO
(i) [\(W \sim \mathrm{Geo}(0.11)\)] \(\mathrm{P}(W = 6) = (0.89)^5(0.11)\)	M1	3.3
\(= 0.06142\ldots\) awrt 0.0614	A1	1.1b
	(2)
(ii) \(\mathrm{P}(W \leqslant 5) = 1 - (0.89)^5\)	M1	3.1b
\(= 0.44159\ldots\) awrt 0.442	A1	1.1b
	(2)
(iii) \(X \sim \mathrm{B}(6, 0.11)\)	M1	3.3
\(\mathrm{P}(X = 4) = 0.001739\ldots\) awrt 0.00174	A1	1.1b
	(2)
(iv) [\(Y \sim \mathrm{NB}(4, 0.11)\)] using a neg bin or \(V \sim \mathrm{B}(6, 0.11)\) and \(\mathrm{P}(V \geqslant 4)\) for M2	M1	3.3
\(\mathrm{P}(Y \leqslant 6) = \mathrm{P}(Y = 4) + \mathrm{P}(Y = 5) + \mathrm{P}(Y = 6)\)	M1	3.1b
\(= (0.11)^4 + \dbinom{4}{3}(0.11)^3(0.89)^1 \times 0.11 + \dbinom{5}{3}(0.11)^3(0.89)^2 \times 0.11\)	M1	3.4
\(= 0.001827\) awrt 0.00183	A1	1.1b
	(4)

Notes

(i) M1 Correct method to find \(\mathrm{P}(W = 6)\) eg \((p)^5(1 - p)\) for \(p = 0.11\) or 0.89

A1 awrt 0.0614 (Correct ans with no incorrect working 2/2)

(ii) M1 Correct method to find \(\mathrm{P}(W \leqslant 5)\)

A1 awrt 0.442 (Correct ans with no incorrect working 2/2)

(iii) M1 For using the model \(\mathrm{B}(6, 0.11)\) allow \(\mathrm{B}(6, 0.89)\) [Implied by 0.0017 or awrt 0.114]

A1 awrt 0.00174 (Correct ans with no incorrect working 2/2)

(iv) In part (iv) we can accept correct expressions or values for probabilities

1^st M1 For using a negative binomial model implied by correct \(\mathrm{P}(Y = 5)\) or \(\mathrm{P}(Y = 6)\)

2^nd M1 Correct method to find \(\mathrm{P}(Y \leqslant 6)\)

3^rd M1 At least two correct terms or \(1 - 0.99817..\) from \(1 - \mathrm{P}(V \leqslant 3)\)

A1 awrt 0.00183

\(a\)	4	5	6
\(\mathrm{P}(Y = a)\)	\(1.46 \times 10^{-4}\)	\(5.21 \times 10^{-4}\)	\(1.16 \times 10^{-3}\)
\(\mathrm{P}(V = a)\)	\(1.74 \times 10^{-3}\)	\(8.60 \times 10^{-5}\)	\(1.77 \times 10^{-6}\)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{P}(\text{Zac wins}) = 0.89 \times 0.11 + (0.89)^3 \times 0.11 + (0.89)^5 \times 0.11 + \ldots\)	M1	3.1b
\(= \dfrac{0.89 \times 0.11}{1 - (0.89)^2}\) oe	M1	1.1b
\(= 0.47089\ldots = 0.471\)*	A1cso*	2.1
	(3)
	Total 13

Notes

1^st M1 Forming the correct probability of Zac winning or identify \(a\) and \(r\) of GP
Allow for \(p = (0.11) \times 0 + (1 - 0.11)(1 - p)\)

2^nd M1 Using sum to infinity of a GP
Allow for \(p = \dfrac{0.89}{1 + 0.89}\)

A1* Previous method marks must be seen leading to an answer 0.471 (NOT awrt 0.471)