Question: [Maximum mark: 6]
Let A and B be two independent events such that P(A ∩ B′) = 0.16 and P(A′ ∩ B) = 0.36.
(a) Given that P(A ∩ B) = x , find the value of x .
(b) Find P(A′ | B′).
▶️Answer/Explanation
Ans:
(a) METHOD 1
EITHER
one of P(A) = x + 0.16 OR P(B) = x + 0.36
OR
THEN
attempt to equate their P(A ∩ B ) with their expression for P (A) × P(B)
P(A ∩ B ) = P (A) × P(B) ⇒x=(x+0.16) ×(x+0.36)
x = 0.24
METHOD 2
attempt to form at least one equation in P(A) and P(B) using independence
(P(A ∩ B’ ) = P (A) × P(B’) ⇒) P(A) × (1-P(B)) = 0.16 OR
(P(A’ ∩ B ) = P (A’) × P(B) ⇒) (1-P(A)) × P(B) = 0.36
P(A) = 0.4 And P(B) = 0.6
P(A ∩ B ) = P (A) × P(B) = 0.4 × 0.6
x = 0.24
(b) METHOD 1
recognising P(A’|B’) = P(A)
= 1- 0.16 – 0.24
= 0.6
METHOD 2
P(B) = 0.36 + 0.24 (=0.6)
\(P(A’|B’)=\frac{P(A’\bigcap B’)}{P(B’)}\left ( =\frac{0.24}{0.4} \right )\)
=0.6
Question
Events \(A\) and \(B\) are such that \({\text{P}}(A) = 0.3\) and \({\text{P}}(B) = 0.4\) .
a.Find the value of \({\text{P}}(A \cup B)\) when
(i) \(A\) and \(B\) are mutually exclusive;
(ii) \(A\) and \(B\) are independent.[4]
b.Given that \({\text{P}}(A \cup B) = 0.6\) , find \({\text{P}}(A|B)\) .[3]
▶️Answer/Explanation
Markscheme
(i) \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) = 0.7\) A1
(ii) \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\) (M1)
\( = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A){\text{P}}(B)\) (M1)
\( = 0.3 + 0.4 – 0.12 = 0.58\) A1
[4 marks]
\({\text{P}}(A \cap B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cup B)\)
\( = 0.3 + 0.4 – 0.6 = 0.1\) A1
\({\text{P}}(A|B) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}}\) (M1)
\( = \frac{{0.1}}{{0.4}} = 0.25\) A1
[3 marks]
Examiners report
Most candidates attempted this question and answered it well. A few misconceptions were identified (eg \({\text{P}}(A \cup B) = {\text{P}}(A){\text{P}}(B)\) ). Many candidates were unsure about the meaning of independent events.
Most candidates attempted this question and answered it well. A few misconceptions were identified (eg \({\text{P}}(A \cup B) = {\text{P}}(A){\text{P}}(B)\) ). Many candidates were unsure about the meaning of independent events.
Question
On Saturday, Alfred and Beatrice play 6 different games against each other. In each game, one of the two wins. The probability that Alfred wins any one of these games is \(\frac{2}{3}\).
Show that the probability that Alfred wins exactly 4 of the games is \(\frac{{80}}{{243}}\).
(i) Explain why the total number of possible outcomes for the results of the 6 games is 64.
(ii) By expanding \({(1 + x)^6}\) and choosing a suitable value for x, prove
\[64 = \left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)\]
(iii) State the meaning of this equality in the context of the 6 games played.
The following day Alfred and Beatrice play the 6 games again. Assume that the probability that Alfred wins any one of these games is still \(\frac{2}{3}\).
(i) Find an expression for the probability Alfred wins 4 games on the first day and 2 on the second day. Give your answer in the form \({\left( {\begin{array}{*{20}{c}}
6 \\
r
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^s}{\left( {\frac{1}{3}} \right)^t}\) where the values of r, s and t are to be found.
(ii) Using your answer to (c) (i) and 6 similar expressions write down the probability that Alfred wins a total of 6 games over the two days as the sum of 7 probabilities.
(iii) Hence prove that \(\left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right) = {\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)^2}\).
Alfred and Beatrice play n games. Let A denote the number of games Alfred wins. The expected value of A can be written as \({\text{E}}(A) = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} \frac{{{a^r}}}{{{b^n}}}\).
(i) Find the values of a and b.
(ii) By differentiating the expansion of \({(1 + x)^n}\), prove that the expected number of games Alfred wins is \(\frac{{2n}}{3}\).
▶️Answer/Explanation
Markscheme
\(B\left( {6,\frac{2}{3}} \right)\) (M1)
\(p(4) = \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right){\left( {\frac{2}{3}} \right)^4}{\left( {\frac{1}{3}} \right)^2}\) A1
\(\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) = 15\) A1
\( = 15 \times \frac{{{2^4}}}{{{3^6}}} = \frac{{80}}{{243}}\) AG
[3 marks]
(i) 2 outcomes for each of the 6 games or \({2^6} = 64\) R1
(ii) \({(1 + x)^6} = \left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)x + \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right){x^2} + \left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right){x^3} + \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right){x^4} + \left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right){x^5} + \left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right){x^6}\) A1
Note: Accept \(^n{C_r}\) notation or \(1 + 6x + 15{x^2} + 20{x^3} + 15{x^4} + 6{x^5} + {x^6}\)
setting x = 1 in both sides of the expression R1
Note: Do not award R1 if the right hand side is not in the correct form.
\(64 = \left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)\) AG
(iii) the total number of outcomes = number of ways Alfred can win no games, plus the number of ways he can win one game etc. R1
[4 marks]
(i) Let \({\text{P}}(x,{\text{ }}y)\) be the probability that Alfred wins x games on the first day and y on the second.
\({\text{P(4, 2)}} = \left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^4} \times {\left( {\frac{1}{3}} \right)^2} \times \left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^2} \times {\left( {\frac{1}{3}} \right)^4}\) M1A1
\({\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\) or \({\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\) A1
r = 2 or 4, s = t = 6
(ii) P(Total = 6) =
P(0, 6) + P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1) + P(6, 0) (M1)
\( = {\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6} + {\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6} + … + {\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\) A2
\( = \frac{{{2^6}}}{{{3^{12}}}}\left( {{{\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)}^2}} \right)\)
Note: Accept any valid sum of 7 probabilities.
(iii) use of \(\left( {\begin{array}{*{20}{c}}
6 \\
i
\end{array}} \right) = \left( {\begin{array}{*{20}{l}}
6 \\
{6 – i}
\end{array}} \right)\) (M1)
(can be used either here or in (c)(ii))
P(wins 6 out of 12) \( = \left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^6} \times {\left( {\frac{1}{3}} \right)^6} = \frac{{{2^6}}}{{{3^{12}}}}\left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right)\) A1
\( = \frac{{{2^6}}}{{{3^{12}}}}\left( {{{\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)}^2}} \right) = \frac{{{2^6}}}{{{3^{12}}}}\left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right)\) A1
therefore \({\left( {\begin{array}{*{20}{c}}
6 \\
0
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
1
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
2
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
3
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
4
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
5
\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}
6 \\
6
\end{array}} \right)^2} = \left( {\begin{array}{*{20}{c}}
{12} \\
6
\end{array}} \right)\) AG
[9 marks]
(i) \({\text{E}}(A) = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} {\left( {\frac{2}{3}} \right)^r}{\left( {\frac{1}{3}} \right)^{n – r}} = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} \frac{{{2^r}}}{{{3^n}}}\)
(a = 2, b = 3) M1A1
Note: M0A0 for a = 2, b = 3 without any method.
(ii) \(n{(1 + x)^{n – 1}} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} r{x^{r – 1}}\) A1A1
(sigma notation not necessary)
(if sigma notation used also allow lower limit to be r = 0)
let x = 2 M1
\(n{3^{n – 1}} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} r{2^{r – 1}}\)
multiply by 2 and divide by \({3^n}\) (M1)
\(\frac{{2n}}{3} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} r\frac{{{2^r}}}{{{3^n}}}\left( { = \sum\limits_{r = 0}^n {\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)} \frac{{{2^r}}}{{{3^n}}}} \right)\) AG
[6 marks]
Examiners report
This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.
(a) Candidates need to be aware how to work out binomial coefficients without a calculator
This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.
(b) (ii) A surprising number of candidates chose to work out the values of all the binomial coefficients (or use Pascal’s triangle) to make a total of 64 rather than simply putting 1 into the left hand side of the expression.
This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.
This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.
(d) This was poorly done. Candidates were not able to manipulate expressions given using sigma notation.
Question
Events \(A\) and \(B\) are such that \({\text{P}}(A) = 0.2\) and \({\text{P}}(B) = 0.5\).
a.Determine the value of \({\text{P}}(A \cup B)\) when
(i) \(A\) and \(B\) are mutually exclusive;
(ii) \(A\) and \(B\) are independent.[4]
b.Determine the range of possible values of \({\text{P}}\left( {A|B} \right)\).[3]
▶️Answer/Explanation
Markscheme
(i) use of \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B)\) (M1)
\({\text{P}}(A \cup B) = 0.2 + 0.5\)
\( = 0.7\) A1
(ii) use of \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A){\text{P}}(B)\) (M1)
\({\text{P}}(A \cup B) = 0.2 + 0.5 – 0.1\)
\( = 0.6\) A1
[4 marks]
\({\text{P}}\left( {A|B} \right) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}}\)
\({\text{P}}\left( {A|B} \right)\) is a maximum when \({\text{P}}(A \cap B) = {\text{P}}(A)\)
\({\text{P}}\left( {A|B} \right)\) is a minimum when \({\text{P}}(A \cap B) = 0\)
\(0 \le {\text{P}}\left( {A|B} \right) \le 0.4\) A1A1A1
Note: A1 for each endpoint and A1 for the correct inequalities.
[3 marks]
Total [7 marks]
Examiners report
This part was generally well done.
Disappointingly, many candidates did not seem to understand the meaning of the word ‘range’ in this context.
Question
A football team, Melchester Rovers are playing a tournament of five matches.
The probabilities that they win, draw or lose a match are \(\frac{1}{2}\), \(\frac{1}{6}\) and \(\frac{1}{3}\) respectively.
These probabilities remain constant; the result of a match is independent of the results of other matches. At the end of the tournament their coach Roy loses his job if they lose three consecutive matches, otherwise he does not lose his job. Find the probability that Roy loses his job.
▶️Answer/Explanation
Markscheme
METHOD 1
to have \(3\) consecutive losses there must be exactly \(5\), \(4\) or \(3\) losses
the probability of exactly \(5\) losses (must be \(3\) consecutive) is \({\left( {\frac{1}{3}} \right)^5}\) A1
the probability of exactly \(4\) losses (with \(3\) consecutive) is \(4{\left( {\frac{1}{3}} \right)^4}\left( {\frac{2}{3}} \right)\) A1A1
Note: First A1 is for the factor \(4\) and second A1 for the other \(2\) factors.
the probability of exactly \(3\) losses (with \(3\) consecutive) is \(3{\left( {\frac{1}{3}} \right)^3}{\left( {\frac{2}{3}} \right)^2}\) A1A1
Note: First A1 is for the factor \(3\) and second A1 for the other \(2\) factors.
(Since the events are mutually exclusive)
the total probability is \(\frac{{1 + 8 + 12}}{{{3^5}}} = \frac{{21}}{{243}}\;\;\;\left( { = \frac{7}{{81}}} \right)\) A1
[6 marks]
METHOD 2
Roy loses his job if
A – first \(3\) games are all lost (so the last \(2\) games can be any result)
B – first \(3\) games are not all lost, but middle \(3\) games are all lost (so the first game is not a loss and the last game can be any result)
or C – first \(3\) games are not all lost, middle \(3\) games are not all lost but last \(3\) games are all lost, (so the first game can be any result but the second game is not a loss)
for A \({4^{{\text{th}}}}\) & \({5^{{\text{th}}}}\) games can be anything
\({\text{P}}(A) = {\left( {\frac{1}{3}} \right)^3} = \frac{1}{{27}}\) A1
for B \({1^{{\text{st}}}}\) game not a loss & \({5^{{\text{th}}}}\) game can be anything (R1)
\({\text{P}}(B) = \frac{2}{3} \times {\left( {\frac{1}{3}} \right)^3} = \frac{2}{{81}}\) A1
for C \({1^{{\text{st}}}}\) game anything, \({2^{{\text{nd}}}}\) game not a loss (R1)
\({\text{P}}(C) = 1 \times \frac{2}{3} \times {\left( {\frac{1}{3}} \right)^3} = \frac{2}{{81}}\) A1
(Since the events are mutually exclusive)
total probability is \(\frac{1}{{27}} + \frac{2}{{81}} + \frac{2}{{81}} = \frac{7}{{81}}\) A1
Note: In both methods all the A marks are independent.
Note: If the candidate misunderstands the question and thinks that it is asking for exactly \(3\) losses award A1 A1 and A1 for an answer of \(\frac{{12}}{{243}}\) as in the last lines of Method 1.
[6 marks]
Total [6 marks]
Examiners report
If a script has lots of numbers with the wrong final answer and no explanation of method it is not going to gain many marks. Working has to be explained. The counting strategy needs to be decided on first. Some candidates misunderstood the context and tried to calculate exactly \(3\) consecutive losses. Not putting a non-loss as \(\frac{2}{3}\) caused unnecessary work.
Question
\(A\) and \(B\) are two events such that \({\text{P}}(A) = 0.25,{\text{ P}}(B) = 0.6\) and \({\text{P}}(A \cup B) = 0.7\).
a.Find \({\text{P}}(A \cap B)\).[2]
b.Determine whether events \(A\) and \(B\) are independent.[2]
▶️Answer/Explanation
Markscheme
\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\)
\({\text{P}}(A \cap B) = 0.25 + 0.6 = 0.7\) M1
\( = 0.15\) A1
[2 marks]
EITHER
\({\text{P}}(A){\text{P}}(B)( = 0.25 \times 0.6) = 0.15\) A1
\( = {\text{P}}(A \cap B)\) so independent R1
OR
\({\text{P}}(A|B) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}} = \frac{{0.15}}{{0.6}} = 0.25\) A1
\( = {\text{P}}(A)\) so independent R1
Note: Allow follow through for incorrect answer to (a) that will result in events being dependent in (b).
[2 marks]
Total [4 marks]
Examiners report
[N/A]
[N/A]