S2 June 2018 Q2

EdexcelOld spec9 marksBinomial Distribution

2. A fair coin is spun 6 times and the random variable \(T\) represents the number of tails obtained.

(a) Give two reasons why a binomial model would be a suitable distribution for modelling \(T\). (2)

(b) Find \(\mathrm{P}(T = 5)\) (2)

A second coin is biased such that the probability of obtaining a head is \(\dfrac{1}{4}\)

This second coin is spun 6 times.

(d) Find the probability that, for the second coin, the number of heads obtained is greater than or equal to the number of tails obtained. (3)

Mark scheme (a)

Scheme	Marks
Only 2 outcomes Heads and Tails oe Constant probability of spinning a Head/Tail oe Coin is spun a fixed number of times oe Each spin of the coin is independent oe	B1 B1
	(2)

Notes

B1 A correct statement – does not need to be in context

B1 A second correct statement in context include coin or heads or tails(do not allow H and T) or spins/flip oe.

Mark scheme (b)

Scheme	Marks
\(T \sim \mathrm{B}(6, 0.5)\)
\(\mathrm{P}(T \leqslant 5) - \mathrm{P}(T \leqslant 4) = 0.9844 - 0.8906\) or \(6\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)\) oe	M1
\(= 0.09375\) or \(\dfrac{3}{32}\) oe awrt 0.0938	A1
	(2)

Notes

M1 [writing or using B(6, 0.5) and writing or using \(\mathrm{P}(T \leqslant 5) - \mathrm{P}(T \leqslant 4)\)] or \(\left[6\left(\dfrac{1}{2}\right)^6 \text{ oe}\right]\)

Mark scheme (c)

Scheme	Marks
\(\mathrm{P}(T = 4,5,6) = 1 - \mathrm{P}(T \leqslant 3)\)	M1
\(= 1 - 0.6563\) \(= 0.3437\) or \(\dfrac{11}{32}\) awrt 0.344	A1
	(2)

Notes

M1 for realising they need find \(\mathrm{P}(T = 4, 5 \text{ or } 6)\) eg \(1 - \mathrm{P}(T \leqslant 3)\) or \(\mathrm{P}(T \geqslant 4)\)

Mark scheme (d)

Scheme	Marks
\(\mathrm{P}(H = 3,4,5,6) = 1 - \mathrm{P}(H \leqslant 2)\)	B1M1d
\(= 1 - 0.8306\) \(= 0.1694\) or \(\dfrac{347}{2048}\) awrt 0.169	A1
	(3)
	(9 marks)

Notes

B1 writing/using B(6, 0.25) and \(\mathrm{P}(H \geqslant 3)\) oe or writing/using B(6, 0.75) and \(\mathrm{P}(T \leqslant 3)\)

M1d dep on B1 for \(1 - \mathrm{P}(H \leqslant 2)\) or dep on B1 \((0.25)^6 + 6(0.75)(0.25)^5 + 15(0.75)^2(0.25)^4 + 20(0.75)^3(0.25)^3\)

A1 awrt 0.169

NB Only accept correct use of H and T in the probability statement unless their variable is correctly defined

NB awrt 0.169 with no incorrect working gains B1M1A1