June 2024 Paper 3 Q1

EdexcelCurrent spec11 marksBinomial Distribution Normal Distribution

1. Xian rolls a fair die 10 times.

The random variable \(X\) represents the number of times the die lands on a six.

(a) Using a suitable distribution for \(X\), find

(i) \(\mathrm{P}(X = 3)\)

(ii) \(\mathrm{P}(X \lt 3)\) (3)

Xian repeats this experiment each day for 60 days and records the number of days when \(X = 3\)

(b) Find the probability that there were at least 12 days when \(X = 3\) (3)

(d) Use a normal approximation to estimate the probability that Xian rolls a total of more than 95 sixes during these 60 days. (4)

Mark scheme (a)

Scheme	Marks	AO
\(X \sim \mathrm{B}\left(10, \tfrac{1}{6}\right)\) [Allow 0.167 or better for \(\tfrac{1}{6}\)]	M1	3.3
(i) [\(\mathrm{P}(X = 3) =\)] 0.155045… awrt 0.155	A1	1.1b
(ii) [\(\mathrm{P}(X \lt 3) = \mathrm{P}(X \leqslant 2) =\)] 0.775226… awrt 0.775	A1	1.1b
	(3)

Notes

If you see any attempt using an \(n\)-sided die with \(n\) not equal to 6 please send to review.

(a) M1 for sight or use of the correct distribution. Must have B, or Bin or Bpd or Bcd and the correct value for \(n\) and \(p\), just \(n = 10, p = \tfrac{1}{6}\) is M0
Implied by one answer correct to 2dp or by sight of \(\dbinom{10}{3}\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)^7\) or one of:
[\(\mathrm{P}(X = 0) =\)] 0.16(1…), [\(\mathrm{P}(X = 1) =\)] 0.32(3…), [\(\mathrm{P}(X = 2) =\)] 0.29(0…), [\(\mathrm{P}(X \leqslant 3) =\)] 0.93(0…)

(i) 1^st A1 for awrt 0.155

(ii) 2^nd A1 for awrt 0.775

Mark scheme (b)

Scheme	Marks	AO
[Let \(D\) = no. of days when \(X = 3\)] \(D \sim \mathrm{B}(60, \text{“}0.155\text{”})\)	M1	3.3
\(\mathrm{P}(D \geqslant 12) = 1 - \mathrm{P}(D \leqslant 11)\) [Allow \(1 - \mathrm{P}(D \lt 12)\)]	M1	3.4
\(= 1 - 0.78819\ldots\) awrt 0.212	A1	1.1b
	(3)

Notes

1^st M1 for selecting a correct model. Sight or use of correct binomial, ft their (a)(i)
May be implied by sight of [\(\mathrm{P}(D \leqslant 11) =\)] 0.78… or 0.79 or [\(\mathrm{P}(D \leqslant 12) =\)] 0.87…

2^nd M1 for correct interpretation of “at least 12” and writing or using \(1 - \mathrm{P}(D \leqslant 11)\)
We are not attempting to ft their incorrect 0.155 on our calculators here.

A1 for awrt 0.212 [Answer only 3/3]

Scheme	Marks	AO
[\(n = 600, p = \tfrac{1}{6}\)] estimate = 100	B1	3.4
	(1)

Mark scheme (d)

Scheme	Marks	AO
[\(S\) = total no. of sixes over 60 days.] \(S \approx T \sim \mathrm{N}\left(\text{``}100\text{''}, \sqrt{\tfrac{5}{6} \times 100}^{\,2}\right)\)	M1A1	3.3, 1.1b
\(\mathrm{P}(S \gt 95) \approx \mathrm{P}([T \gt]\,95.5)\) or \(\mathrm{P}\left([Z \gt]\,\dfrac{95.5 - \text{``}100\text{''}}{\text{``}9.128\ldots\text{''}}\right)\) or \(\mathrm{P}([Z \gt]\,{-0.49}..)\)	M1	3.4
\(= 0.688976\ldots\) awrt 0.689	A1	1.1b
	(4)
	(11 marks)

Notes

1^st M1 for attempting normal with mean = 100 or ft their answer to (c)
May be implied by the correct mean and a correctly labelled s.d. (\(\sigma\)) or var (\(\sigma^2\))

1^st A1 for correctly labelled standard deviation allow \(\sqrt{\tfrac{250}{3}} = \sqrt{83.3\ldots} = 9.1(28\ldots)\) or correctly labelled variance. Implied by \(\mathrm{N}\left(\mu, \tfrac{250}{3}\right)\) or correct answer.

2^nd M1 for attempt at continuity correction i.e. sight of \(95 \pm 0.5\)

2^nd A1 for awrt 0.689 [Answer only 4/4]

NB If they don’t state the model for 1^st M1 but just give probabilities with probability statements (\(Y\) is any letter):

\(\sigma = \tfrac{250}{3}\): 1^st M1 implied by: \(\mathrm{P}(Y \gt 94.5) = 0.52(63\ldots)\), \(\mathrm{P}(Y \gt 95) = 0.52(39\ldots)\), \(\mathrm{P}(Y \gt 95.5) = 0.52(15..)\)

No cc: 1^st M1 1^st A1 implied by: \(\mathrm{P}(T \gt 95) = 0.70(805\ldots)\)
1^st M1 1^st A1 2^nd M1 implied by: \(\mathrm{P}(T \gt 94.5) = 0.72(657\ldots)\)

Exact binomial gives 0.68567… and will likely score 0/4