One unknown from sum constraint only - Discrete Probability Distributions

Edexcel S1 2004 November Q4

4. The discrete random variable $X$ has probability function $$\mathrm { P } ( X = x ) = \begin{array} { l l } 0.2 , & x = - 3 , - 2
\alpha , & x = - 1,0
0.1 , & x = 1,2 . \end{array}$$ Find

$\alpha$,
$\mathrm { P } ( - 1 \leq X < 2 )$,
$\mathrm { F } ( 0.6 )$,
the value of $a$ such that $\mathrm { E } ( a X + 3 ) = 1.2$,
$\operatorname { Var } ( X )$,
$\operatorname { Var } ( 3 X - 2 )$.

Edexcel S1 Q4

4. The discrete random variable $X$ has the following probability distribution :

$x$	0	1	2	3	4	5
$\mathrm { P } ( X = x )$	0.11	0.17	0.2	0.13	$p$	$p ^ { 2 }$

Find the value of $p$.
Find
1. $\mathrm { P } ( 0 < X \leq 2 )$,
2. $\mathrm { P } ( X \geq 3 )$.
Find the mean and the variance of $X$.
Construct a table to represent the cumulative distribution function $\mathrm { F } ( x )$.

Edexcel S1 Q4

4. The discrete random variable $X$ has the following probability distribution.

$x$	1	2	3	4	5
$\mathrm { P } ( X = x )$	0.1	0.35	$k$	0.15	$k$

Calculate

$k$,
$\mathrm { F } ( 2 )$,
$\mathrm { P } ( 1.3 < X \leq 3.8 )$,
$\mathrm { E } ( X )$,
$\operatorname { Var } ( 3 X + 2 )$.

Edexcel S1 Q1

The discrete random variable $X$ has the following probability distribution.

$x$	$k$	$k + 4$	$2 k$
$\mathrm { P } ( X = x )$	$\frac { 1 } { 8 }$	$\frac { 3 } { 8 }$	$\frac { 1 } { 2 }$

Find and simplify an expression in terms of $k$ for $\mathrm { E } ( X )$. Given that $\mathrm { E } ( X ) = 9$,
find the value of $k$.

AQA S2 2007 January Q4

4 The number of fish, $X$, caught by Pearl when she goes fishing can be modelled by the following discrete probability distribution:

$\boldsymbol { x }$	1	2	3	4	5	6	$\geqslant 7$
$\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )$	0.01	0.05	0.14	0.30	$k$	0.12	0

Find the value of $k$.
Find:
1. $\mathrm { E } ( X )$;
2. $\operatorname { Var } ( X )$.
When Pearl sells her fish, she earns a profit, in pounds, given by $$Y = 5 X + 2$$ Find:
1. $\mathrm { E } ( Y )$;
2. the standard deviation of $Y$.

OCR MEI Further Statistics A AS 2024 June Q1

1 The probability distribution for a discrete random variable $X$ is given in the table below.

$x$	0	1	2	3
$\mathrm { P } ( \mathrm { X } = \mathrm { x } )$	$2 c$	$3 c$	$0.5 - c$	$c$

Find the value of $c$.
Find the value of each of the following.
- $\mathrm { E } ( X )$
- $\operatorname { Var } ( X )$
The random variable $Y$ is defined by $Y = 2 X - 3$.
Find the value of each of the following.
- E(Y)
- $\operatorname { Var } ( Y )$

WJEC Further Unit 2 2022 June Q1

The probability distribution for the prize money, $\pounds X$ per ticket, in a local fundraising lottery is shown below.

$x$	0	2	100	1000
$\mathrm { P } ( X = x )$	0.9	0.09	$p$	0.0001

Calculate the value of $p$.
Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
1. What is the minimum lottery ticket price that the organiser should set in order to make a profit in the long run?
2. Suggest why, in practice, people would be prepared to pay more than this minimum price.

OCR FS1 AS 2021 June Q2

2 The probability distribution for the discrete random variable $W$ is given in the table.

$w$	1	2	3	4
$\mathrm { P } ( W = w )$	0.25	0.36	$x$	$x ^ { 2 }$

Show that $\operatorname { Var } ( W ) = 0.8571$.
Find $\operatorname { Var } ( 3 W + 6 )$. In this question you must show detailed reasoning.
The random variable $T$ has a binomial distribution. It is known that $\mathrm { E } ( T ) = 5.625$ and the standard deviation of $T$ is 1.875 . Find the values of the parameters of the distribution.

AQA AS Paper 2 2018 June Q13

13 The table below shows the probability distribution for a discrete random variable $X$.

$\boldsymbol { x }$	0	1	2	3	4 or more
$\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )$	0.35	0.25	$k$	0.14	0.1

Find the value of $k$. Circle your answer.
0.140 .160 .1801

AQA AS Paper 2 2024 June Q15

3 marks

15 The number of flowers which grow on a certain type of plant can be modelled by the discrete random variable $X$ The probability distribution of $X$ is given in the table below.

$\boldsymbol { x }$	0	1	2	3	4	5 or more
$\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )$	0.03	0.15	0.22	0.31	0.09	$p$

15

Find the value of $p$
15
Two plants of this type are randomly selected from a large batch received from a local garden centre. Find the probability that the two plants will produce a total of three flowers.
[0pt] [3 marks]
15
1. State one assumption necessary for the calculation in part (b) to be valid. 15
(ii) Comment on whether, in reality, this assumption is likely to be valid.

\(x\)	1	2	3	4	5
\(\mathrm { P } ( X = x )\)	0.1	0.35	\(k\)	0.15	\(k\)

\(x\)	\(k\)	\(k + 4\)	\(2 k\)
\(\mathrm { P } ( X = x )\)	\(\frac { 1 } { 8 }\)	\(\frac { 3 } { 8 }\)	\(\frac { 1 } { 2 }\)

\(x\)	0	1	2	3
\(\mathrm { P } ( \mathrm { X } = \mathrm { x } )\)	\(2 c\)	\(3 c\)	\(0.5 - c\)	\(c\)

\(w\)	1	2	3	4
\(\mathrm { P } ( W = w )\)	0.25	0.36	\(x\)	\(x ^ { 2 }\)

\(\boldsymbol { x }\)	1	2	3	4	5	6	\(\geqslant 7\)
\(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\)	0.01	0.05	0.14	0.30	\(k\)	0.12	0