A2 June 2023 Q7

EdexcelCurrent spec13 marksGeometric & Negative Binomial

7. Each time a spinner is spun, the probability that it lands on red is 0.2

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

(i) for the 1st time on the 4th spin (2)

(ii) for the 3rd time on the 8th spin (2)

(iii) exactly 4 times during 10 spins (2)

Each time the spinner is spun, the probability that it lands on yellow is 0.4

In a game with this spinner, a player must choose one of two events

\(R\) is the event that the spinner lands on red for the 1st time in at most 4 spins
\(Y\) is the event that the spinner lands on yellow for the 3rd time in at most 7 spins

(b) Showing your calculations clearly, determine which of these events has the greater probability. (7)

Mark scheme (a)

Scheme	Marks	AO
(i) \(X \sim \mathrm{Geo}(0.2)\) or \(\mathrm{P}(X = 4) = 0.8^3 \times 0.2\)	M1	3.3
\(= \underline{\mathbf{0.1024}}\)	A1	1.1b
	(2)
(ii) \(T \sim \mathrm{NegBin}(3, 0.2)\) or \(\mathrm{P}(T = 8) = \dbinom{7}{2}0.2^2 \times 0.8^5 \times 0.2\)	M1	3.3
\(= 0.05505\ldots\) awrt 0.0551	A1	1.1b
	(2)
(iii) \(F \sim \mathrm{B}(10, 0.2)\) or \(\mathrm{P}(F = 4) = \dbinom{10}{4}0.2^4 \times 0.8^6\)	M1	3.3
\(\mathrm{P}(F = 4) = 0.088080\ldots\) awrt 0.0881	A1	1.1b
	(2)

Notes

(i) M1 for selecting the correct model. Stated or used which may be implied by ans.

A1 for 0.1024 or \(\frac{64}{625}\) (accept 0.102) (correct answer scores 2 out of 2)

(ii) M1 for selecting the correct model. Stated or used which may be implied by ans.
Allow \(0.2 \times \mathrm{P}(V = 2)\) from \(V \sim \mathrm{B}(7, 0.2)\)

A1 for awrt 0.0551 (correct answer scores 2 out of 2)

(iii) M1 for selecting the correct model. Stated or used may be implied by ans of 0.967(2)

A1 for awrt 0.0881 (correct answer scores 2 out of 2)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{P}(R) = \mathrm{P}(X \leqslant 4)\) and \(X \sim \mathrm{Geo}(0.2)\) or \(\mathrm{P}(X \geqslant 1),\ X \sim \mathrm{B}(4, 0.2)\)	M1	3.1b
\(= 1 - \mathrm{P}(X \gt 4) = 1 - 0.8^4\) or \(= 1 - \mathrm{P}(X = 0) = 1 - 0.8^4\)	M1	3.4
\(= \underline{0.59(04)}\)	A1	1.1b
\(\mathrm{P}(Y) = \mathrm{P}(N \leqslant 7)\) and \(N \sim \mathrm{NegBin}(3, 0.4)\)	M1	3.1b
\(0.4^3 + \dbinom{3}{2}0.4^3 0.6^1 + \dbinom{4}{2}0.4^3 0.6^2 + \dbinom{5}{2}0.4^3 0.6^3 + \dbinom{6}{2}0.4^3 0.6^4\) or \(1 - \left(\dbinom{7}{2}0.4^2 0.6^5 + \dbinom{7}{1}0.4^1 0.6^6 + \dbinom{7}{0}0.6^7\right)\)	M1	3.4
ALT: \(\mathrm{P}(Y) = \mathrm{P}(W \gt 2)\) where \(W \sim \mathrm{B}(7, 0.4)\) [M1] \(= 1 - \mathrm{P}(W \leqslant 2)\ [= 1 - 0.419904]\) [M1]
\(= \underline{0.58(0096)}\)	A1	1.1b
\(R\) (has the greater probability)	A1	3.2b
	(7)
	(13 marks)

Notes

1^st M1 for a correct distribution and prob. expression for \(\mathrm{P}(R)\) (may be implied by 2^nd M1)

2^nd M1 for a correct numerical expression for \(\mathrm{P}(R)\) (allow any equivalent expression) (corrected from the printed mark scheme: the binomial alternative printed \(1 - \mathrm{P}(Y = 0)\); with \(X \sim \mathrm{B}(4, 0.2)\) it is \(1 - \mathrm{P}(X = 0)\))

1^st A1 for awrt 0.590 or \(\frac{369}{625}\) (accept 0.59 or better) awrt 0.590 implies M1M1A1

3^rd M1 for a correct distribution and prob. expression for \(\mathrm{P}(Y)\) (may be implied by 4^th M1)

4^th M1 for a correct numerical expression for \(\mathrm{P}(Y)\) (allow any equivalent expression)

2^nd A1 for awrt 0.580 or (accept 0.58 or better) awrt 0.580 implies M1M1A1

3^rd A1 dep on all other marks for \(R\) or correct description in words Condone \(\mathrm{P}(R) \gt \mathrm{P}(Y)\)