Question
A biased four-sided die, A, is rolled. Let X be the score obtained when die A is rolled. The
probability distribution for X is given in the following table.
(a) Find the value of p . [2]
(b) Hence, find the value of E (X ) . [2]
A second biased four-sided die, B, is rolled. Let Y be the score obtained when die B is rolled.
The probability distribution for Y is given in the following table.
(c) (i) State the range of possible values of r .
(ii) Hence, find the range of possible values of q . [3]
(d) Hence, find the range of possible values for E (Y ) . [3]
Agnes and Barbara play a game using these dice. Agnes rolls die A once and Barbara rolls die B once. The probability that Agnes’ score is less than Barbara’s score is \(\frac{1}{2}\)
(e) Find the value of E (Y ) . [6]
Answer/Explanation
Ans
Question
A four-sided die has three blue faces and one red face. The die is rolled.
Let B be the event a blue face lands down, and R be the event a red face lands down.
Write down
(i) P(B);
(ii) P(R).
If the blue face lands down, the die is not rolled again. If the red face lands down, the die is rolled once again. This is represented by the following tree diagram, where p, s, t are probabilities.
Find the value of p, of s and of t.
Guiseppi plays a game where he rolls the die. If a blue face lands down, he scores 2 and is finished. If the red face lands down, he scores 1 and rolls one more time. Let X be the total score obtained.
(i) Show that \({\text{P}}(X = 3) = \frac{3}{{16}}\) .
(ii) Find \({\text{P}}(X = 2)\) .
(i) Construct a probability distribution table for X.
(ii) Calculate the expected value of X.
If the total score is 3, Guiseppi wins \(\$ 10\). If the total score is 2, Guiseppi gets nothing.
Guiseppi plays the game twice. Find the probability that he wins exactly \(\$ 10\).
Answer/Explanation
Markscheme
(i) P(B) \( = \frac{3}{4}\) A1 N1
(ii) P(R) \( = \frac{1}{4}\) A1 N1
[2 marks]
\(p = \frac{3}{4}\) A1 N1
\(s = \frac{1}{4}\), \(t = \frac{3}{4}\) A1 N1
[2 marks]
(i) \({\text{P}}(X = 3)\)
\( = {\text{P (getting 1 and 2)}} = \frac{1}{4} \times \frac{3}{4}\) A1
\( = \frac{3}{{16}}\) AG N0
(ii) \({\text{P}}(X = 2) = \frac{1}{4} \times \frac{1}{4} + \frac{3}{4}{\text{ }}\left( {{\text{or }}1 – \frac{3}{{16}}} \right)\) (A1)
\( = \frac{{13}}{{16}}\) A1 N2
[3 marks]
(i)
A2 N2
(ii) evidence of using \({\text{E}}(X) = \sum{x{\text{P}}(X = x)} \) (M1)
\({\text{E}}(X) = 2\left( {\frac{{13}}{{16}}} \right) + 3\left( {\frac{3}{{16}}} \right)\) (A1)
\( = \frac{{35}}{{16}}{\text{ }}\left( { = 2\frac{3}{{16}}} \right)\) A1 N2
[5 marks]
win \(\$ 10 \Rightarrow \) scores 3 one time, 2 other time (M1)
\({\text{P}}(3) \times {\text{P}}(2) = \frac{{13}}{{16}} \times \frac{3}{{16}}\) (seen anywhere) A1
evidence of recognising there are different ways of winning \(\$ 10\) (M1)
e.g. \({\text{P}}(3) \times {\text{P}}(2) + {\text{P}}(2) \times {\text{P}}(3)\) , \(2\left( {\frac{{13}}{{16}} \times \frac{3}{{16}}} \right)\) , \(\frac{{36}}{{256}} + \frac{3}{{256}} + \frac{{36}}{{256}} + \frac{3}{{256}}\)
\({\text{P(win }}\$ 10) = \frac{{78}}{{256}}{\text{ }}\left( { = \frac{{39}}{{128}}} \right)\) A1 N3
[4 marks]
Question
José travels to school on a bus. On any day, the probability that José will miss the bus is \(\frac{1}{3}\) .
If he misses his bus, the probability that he will be late for school is \(\frac{7}{8}\) .
If he does not miss his bus, the probability that he will be late is \(\frac{3}{8}\) .
Let E be the event “he misses his bus” and F the event “he is late for school”.
The information above is shown on the following tree diagram.
Find
(i) \({\rm{P}}(E \cap F)\) ;
(ii) \({\rm{P}}(F)\) .
Find the probability that
(i) José misses his bus and is not late for school;
(ii) José missed his bus, given that he is late for school.
The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.
Copy and complete the probability distribution table.
The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.
Find the expected cost for José for both days.
Answer/Explanation
Markscheme
(i) \(\frac{7}{{24}}\) A1 N1
(ii) evidence of multiplying along the branches (M1)
e.g. \(\frac{2}{3} \times \frac{5}{8}\) , \(\frac{1}{3} \times \frac{7}{8}\)
adding probabilities of two mutually exclusive paths (M1)
e.g. \(\left( {\frac{1}{3} \times \frac{7}{8}} \right) + \left( {\frac{2}{3} \times \frac{3}{8}} \right)\) , \(\left( {\frac{1}{3} \times \frac{1}{8}} \right) + \left( {\frac{2}{3} \times \frac{5}{8}} \right)\)
\({\rm{P}}(F) = \frac{{13}}{{24}}\) A1 N2
[4 marks]
(i) \(\frac{1}{3} \times \frac{1}{8}\) (A1)
\(\frac{1}{{24}}\) A1
(ii) recognizing this is \({\rm{P}}(E|F)\) (M1)
e.g. \(\frac{7}{{24}} \div \frac{{13}}{{24}}\)
\(\frac{{168}}{{312}}\) \(\left( { = \frac{7}{{13}}} \right)\) A2 N3
[5 marks]
A2A1 N3
[3 marks]
correct substitution into \({\rm{E}}(X)\) formula (M1)
e.g. \(0 \times \frac{1}{9} + 3 \times \frac{4}{9} + 6 \times \frac{4}{9}\) , \(\frac{{12}}{9} + \frac{{24}}{9}\)
\({\rm{E}}(X) = 4\) (euros) A1 N2
[2 marks]
Question
The probability distribution of a discrete random variable X is given by \[{\rm{P}}(X = x) = \frac{{{x^2}}}{{14}}{\text{, }}x \in \left\{ {1{\text{, }}2{\text{, }}k} \right\}{\text{, where}} k > 0\] .
Write down \({\rm{P}}(X = 2)\) .
Show that \(k = 3\) .
Find \({\rm{E}}(X)\) .
Answer/Explanation
Markscheme
\({\rm{P}}(X = 2) = \frac{4}{{14}}\) \(\left( { = \frac{2}{7}} \right)\) A1 N1
[1 mark]
\({\rm{P}}(X = 1) = \frac{1}{{14}}\) (A1)
\({\rm{P}}(X = k) = \frac{{{k^2}}}{{14}}\) (A1)
setting the sum of probabilities \( = 1\) M1
e.g. \(\frac{1}{{14}} + \frac{4}{{14}} + \frac{{{k^2}}}{{14}} = 1\) , \(5 + {k^2} = 14\)
\({k^2} = 9\) (accept \(\frac{{{k^2}}}{{14}} = \frac{9}{{14}}\) ) A1
\(k = 3\) AG N0
[4 marks]
correct substitution into \({\rm{E}}(X) = \sum {x{\rm{P}}(X = x)} \) A1
e.g. \(1\left( {\frac{1}{{14}}} \right) + 2\left( {\frac{4}{{14}}} \right) + 3\left( {\frac{9}{{14}}} \right)\)
\({\rm{E}}(X) = \frac{{36}}{{14}}\) \(\left( { = \frac{{18}}{7}} \right)\) A1 N1
[2 marks]
Question
The random variable X has the following probability distribution.
Given that \({\rm{E}}(X) = 1.7\) , find q .
Answer/Explanation
Markscheme
correct substitution into \({\rm{E}}(X) = \sum {px} \) (seen anywhere) A1
e.g. \(1s + 2 \times 0.3 + 3q = 1.7\) , \(s + 3q = 1.1\)
recognizing \(\sum {p = 1} \) (seen anywhere) (M1)
correct substitution into \(\sum {p = 1} \) A1
e.g. \(s + 0.3 + q = 1\)
attempt to solve simultaneous equations (M1)
correct working (A1)
e.g. \(0.3 + 2q = 0.7\) , \(2s = 1\)
\(q = 0.2\) A1 N4
[6 marks]