S2 June 2014 (R) Q3

EdexcelOld spec10 marksPoisson Distribution

3. Accidents occur randomly at a road junction at a rate of 18 every year.
The random variable \(X\) represents the number of accidents at this road junction in the next 6 months.

(a) Write down the distribution of \(X\). (2)

(b) Find \(\mathrm{P}(X \gt 7)\). (2)

(c) Show that the probability of at least one accident in a randomly selected month is 0.777 (correct to 3 decimal places). (3)

(d) Find the probability that there is at least one accident in exactly 4 of the next 6 months. (3)

Scheme	Marks
\(X \sim \mathrm{Po}(9)\)	M1A1
	(2)

Mark scheme (b)

Scheme	Marks
\(\mathrm{P}(X \gt 7) = 1 - \mathrm{P}(X \leqslant 7)\)	M1
\(= \left[1 - 0.3239\right] = 0.6761\)	A1
	(2)

Notes

M1 for writing \(1 - \mathrm{P}(X \leqslant 7)\). This may be implied by \(1 - 0.3239\) or a correct answer
A1 for awrt 0.676

Mark scheme (c)

Scheme	Marks
[\(Y\) = no. of accidents in a month] \(Y \sim \mathrm{Po}(1.5)\)	B1
\(\mathrm{P}(Y \geqslant 1) = 1 - \mathrm{P}(Y = 0)\)	M1
\(= \left[1 - 0.2231\right] = 0.7769\) (= 0.777 (3dp))*	A1cso
	(3)

Notes

B1 Po(1.5) written or used
M1 writing or using \(1 - \mathrm{P}(Y = 0)\) or \(1 - \mathrm{P}(Y \leqslant 0)\) or \(1 - \mathrm{e}^{-\lambda}\) [may not be \(Y\)]
A1 for at least (1 – 0.223) or better. No need for final comment. * answer given so 0.777 does not imply all three marks

Mark scheme (d)

Scheme	Marks
[\(A\) = no. of months with at least one accident] \(A \sim \mathrm{B}(6, 0.777)\)	M1
\(\mathrm{P}(A = 4) = \dbinom{6}{4}(0.777)^4(0.223)^2\)	M1
\(= 0.2719\ldots\) awrt 0.272	A1
	(3)
	(10 marks)

Notes

1st M1 for identifying binomial with \(n = 6\) and \(p = 0.777\) or better. Condone use of \(p = 0.223\). May be implied by \((p)^4(1 - p)^2\), \(p\) = awrt 0.777 or awrt 0.223

2nd M1 Must have \({}^6\mathrm{C}_4\,(0.777)^4(1 - 0.777)^2\)

A1 for awrt 0.272