The discrete random variable R is such that R ∈ 1 , 3 , 5 , 7 Given that P R = r = k r + 1 r + 2 , find the value of the constant k .

Given that, R ∈ { 1 , 3 , 5 , 7 } ⇒ R = 1 , P ( R = 1 ) = k × 2 3 = 2 k 3 ⇒ R = 3 , P ( R = 3 ) = k × 4 5 = 4 k 5 ⇒ R = 5 , P ( R = 5 ) = k × 6 7 = 6 k 7 ⇒ R = 7 , P ( R = 7 ) = k × 8 9 = 8 k 9 Now, P ( R = 1 ) + P ( R = 3 ) + P ( R = 5 ) + P ( R = 7 ) = 1 ⇒ 2 k 3 + 4 k 5 + 6 k 7 + 8 k 9 = 1 ⇒ k = 315 1012 The value of constant k is 315 1012 .

The discrete random variable R is such that R ∈ 1 , 3 , 5 , 7 Given that P R = r = k r + 1 r + 2 , find the value of the constant k .

Given that, R ∈ { 1 , 3 , 5 , 7 } ⇒ R = 1 , P ( R = 1 ) = k × 2 3 = 2 k 3 ⇒ R = 3 , P ( R = 3 ) = k × 4 5 = 4 k 5 ⇒ R = 5 , P ( R = 5 ) = k × 6 7 = 6 k 7 ⇒ R = 7 , P ( R = 7 ) = k × 8 9 = 8 k 9 Now, P ( R = 1 ) + P ( R = 3 ) + P ( R = 5 ) + P ( R = 7 ) = 1 ⇒ 2 k 3 + 4 k 5 + 6 k 7 + 8 k 9 = 1 ⇒ k = 315 1012 The value of constant k is 315 1012 .

The discrete random variable R is such that R ∈ 1 , 3 , 5 , 7 Given that P R = r = k r + 1 r + 2 , find the value of the constant k . Hence, find P R ⩽ 4 .

Given that, R ∈ { 1 , 3 , 5 , 7 } ⇒ R = 1 , P ( r = 1 ) = k × 2 3 = 2 k 3 ⇒ R = 3 , P ( r = 3 ) = k × 4 5 = 4 k 5 ⇒ R = 5 , P ( r = 5 ) = k × 6 7 = 6 k 7 ⇒ R = 7 , P ( r = 7 ) = k × 8 9 = 8 k 9 Now, P ( r = 1 ) + P ( r = 3 ) + P ( r = 5 ) + P ( r = 7 ) = 1 ⇒ 2 k 3 + 4 k 5 + 6 k 7 + 8 k 9 = 1 ∴ 210 k + 252 k + 270 k + 280 k 315 = 1 ⇒ k = 315 1012 ⇒ P ( R ⩽ 4 ) = P ( r = 1 ) + P ( r = 3 ) = 2 k 3 + 4 k 5 = 22 k 15 = 22 15 × 315 1012 ∴ P ( R ⩽ 4 ) = 21 46

The probability distribution for the random variable X is given in the following table. y 0 1 2 3 P X = x 0 . 10 0 . 12 0 . 36 0 . 42 Calculate E X and Var X

Given that, The Probability distribution for the random variable X is given in the following table, x 0 1 2 3 P X = x 0 . 1 0 . 12 0 . 36 0 . 42 ∴ E ( x ) = ∑ i = 1 n x i × P i = x 1 × P 1 + x 2 × P 2 + x 3 × P 3 + x 4 × P 4 = 0 × 0 . 1 + 1 × 0 . 12 + 2 × 0 . 36 + 3 × 0 . 42 = 0 + 0 . 12 + 0 . 72 + 1 . 26 ∴ E ( x ) = 2 · 1 Now, Var ( x ) = E x 2 - [ E ( x ) ] 2 Va r ( x ) = ∑ i = 1 n x i P i 2 - ∑ i = 1 n x i P i 2 ∑ l = 1 n x i 2 P i = ( 0 ) 2 × 0 . 1 + ( 1 ) 2 × 0 . 12 + ( 2 ) 2 × 0 . 36 + ( 3 ) 2 × 0 . 42 = 0 + 0 . 12 + 4 × 0 . 36 + 9 × 0 . 42 ⇒ ∑ l = 1 n x i 2 P i = 5 . 34 from (1) V ar ( x ) = 5 . 34 - ( 2 · 1 ) 2 = 5 . 34 - 4 . 41 ⇒ V ar ( x ) = 0 . 93

The probability distribution for the random variable Y is given in the following table. y 0 1 2 3 4 P Y = y 0 . 03 2 p 0 . 32 p 0 . 05 Find the value of p .

Given that, The probability distribution for the random variable Y is given in the following table. y 0 1 2 3 4 P Y = y 0 . 03 2 p 0 . 32 p 0 . 05 We know that, ⇒ P ( y = 0 ) + P ( Y = 1 ) + p ( 4 = 2 ) + P ( y = 3 ) + P ( y = 4 ) = 1 ⇒ 0 . 03 + 2 p + 0 . 32 + P + 0 . 05 = 1 ⇒ 3 p = 0 . 6 ∴ p = 0 . 2

The probability distribution for the random variable Y is given in the following table. y 0 1 2 3 4 P Y = y 0 . 03 2 p 0 . 32 p 0 . 05 Calculate E Y and the standard deviation of Y .

Given that, The probability distribution for the random variable Y is given in the following table. y 0 1 2 3 4 P Y = y 0 . 03 2 p 0 . 32 p 0 . 05 a) We know that, ⇒ P ( y = 0 ) + P ( Y = 1 ) + p ( 4 = 2 ) + P ( y = 3 ) + P ( y = 4 ) = 1 ⇒ 0 . 03 + 2 p + 0 . 32 + P + 0 . 05 = 1 ⇒ 3 p = 0 . 6 ∴ p = 0 . 2 y p y 2 p y p y 2 0 0 . 03 0 0 0 1 2 p 1 2 p 2 p 2 0 . 32 4 0 . 64 1 . 28 3 p 9 3 p 9 p 4 0 . 05 16 0 . 2 0 . 8 ∑ p y = 1 . 84 ∑ p y 2 = 4 . 28 E ( x ) = ∑ i = 1 n p i y i = P 1 × y 1 + P 2 × y 2 + P 3 × y 3 + P 4 × y 4 + P 5 × y 5 ⇒ E ( x ) = 1 . 84 Var ( x ) = ∑ i = 1 n p i y i 2 - ∑ E = 1 n p i y i 2 = 4 . 28 - ( 1 . 84 ) 2 ⇒ Var ( x ) = 0 . 895 Standard deviation = V a r ( x ) = 0 . 895 ∴ Standard deviation = 0 . 946

The random variable T is such that T ∈ 1 , 3 , 6 , 10 Given that the four possible values of T are equiprobable, find E ( T ) and Var T

Given that, The random variable T is such that T ∈ 1 , 3 , 6 , 10 Given that the four possible values of T are equiprobable, P ( T = 1 ) = 1 4 , P ( T = 3 ) = 1 4 , P ( T = 6 ) = 1 4 , P ( T = 10 ) = 1 4 T P P T P T 2 1 1 4 1 4 1 4 3 1 4 3 4 9 4 6 1 4 3 2 36 4 10 1 4 5 2 100 4 ∑ P T = 20 4 = 5 ∑ P T 2 = 146 4 = 36 . 5 ∴ E ( T ) = ∑ i = 1 n P i T i = P 1 T 1 + P 2 T 2 + P 3 T 3 + P 4 T 4 ∴ E ( T ) = 20 4 = 5 ∴ V a r ( T ) = ∑ i = 1 n P i T i 2 - [ E ( T ) ] 2 = 36 . 5 - ( 5 ) 2 ⇒ Var ( T ) = 11 . 5

The following table shows the probability distribution for the random variable V . y 1 3 9 m P V = v 0 . 4 0 . 28 0 . 14 0 . 18 Given that E V = 5 . 38 , find the value of m and calculate Var V .

Given that, y 1 3 9 m P V = v 0 . 4 0 . 28 0 . 14 0 . 18 We know that, E ( V ) = ∑ i = 1 n P i v i = P 1 V 1 + P 2 V 2 + P 3 V 3 + P 4 V 4 ⇒ 5 . 38 = ( 1 × 0 . 4 ) + ( 3 × 0 . 28 ) + ( 9 × 0 . 16 ) + ( m × 0 . 18 ) ∴ 5 . 38 = 0 . 6 + 0 . 84 + 1 . 26 + 0 . 18 m ∴ 2 . 88 = 0 . 18 m ∴ m = 16 ⇒ Var ( v ) = ∑ i = 1 n P i v i 2 - ( E ( v ) ) 2 = ( 1 ) 2 × 0 . 4 + ( 3 ) 2 × 0 . 28 + ( 9 ) 2 × 0 . 14 + ( 16 ) 2 × 0 . 13 - ( 5 . 38 ) 2 = [ 0 . 4 + 2 . 52 + 11 . 34 + 46 . 08 ] - 28 . 944 = 60 . 34 - 28 . 944 ⇒ Var ( v ) = 31 . 396

R is a random variable such that R ∈ 10 , 20 , 70 , 100 . Given that P ( R = r ) is proportional to r , show that E R = 77 and find Var R

Given that, r 10 20 70 100 P R = r 0 . 05 0 . 1 0 . 35 0 . 5 We know that, E R = ∑ i = 1 n r i × P i ⇒ E ( R ) = 10 × 0 . 05 + 20 × 0 . 1 + 70 × 0 . 35 + 100 × 0 . 5 ⇒ E ( R ) = 77 V a r R = ∑ i = 1 n r 2 i × P i ⇒ Var ( R ) = 10 2 × 0 . 05 + 20 2 × 0 . 1 + 70 2 × 0 . 35 + 100 2 × 0 . 5 - 77 2 ⇒ Var ( R ) = 831

The probability distribution for the random variable W is given in the following table. w 2 7 a 24 P W = w 0 . 3 0 . 3 0 . 1 0 . 3 Given that E W = a find a and evaluate Var W

The probability distribution for the random variable W is given in the following table. w 2 7 a 24 P W = w 0 . 3 0 . 3 0 . 1 0 . 3 Given that E W = a find a and evaluate Var W w P W = w P w P w 2 2 0 . 3 0 . 6 1 . 2 7 0 . 3 2 . 1 14 . 7 a 0 . 1 0 . 1 a 12 . 1 24 0 . 3 7 . 2 172 . 8 9 . 9 + 0 . 1 a 200 . 8 But, E ( w ) = ∑ i = 1 n P i w i = 9 · 9 + 0 . 1 a ∴ a = 9 . 9 + 0 . 1 a ∴ 0 . 9 a = 9 . 9 ∴ a = 11 ⇒ V ar ( W ) = ∑ i = 1 n p i w i 2 - E ( w ) 2 -------------- ( ∴ E ( ω ) = a = 11 ) = 200 . 8 - ( 11 ) 2 = 200 . 8 - 121 ⇒ V ar ( W ) = 79 . 8

EASY

AS and A Level

IMPORTANT

Earn 100

The discrete random variable $R$ is such that $R \in \{1, 3, 5, 7\}$

Given that $P (R = r) = \frac{k (r + 1)}{r + 2},$ find the value of the constant $k .$

Important Questions on Probability Distributions

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 6 - Probability Distributions>EXERCISE 6A>Q 18

The discrete random variable $R$ is such that $\Rightarrow k = \frac{315}{1012}$

Given that $P (R = r) = \frac{k (r + 1)}{r + 2},$ find the value of the constant $k .$ Hence, find $P (R ⩽ 4) .$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 6 - Probability Distributions>EXERCISE 6B>Q 1

The probability distribution for the random variable $X$ is given in the following table.

$y$	$0$	$1$	$2$	$3$
$P (X = x)$	$0.10$	$0.12$	$0.36$	$0.42$

Calculate $E (X)$ and $Var (X)$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 6 - Probability Distributions>EXERCISE 6B>Q 2

The probability distribution for the random variable $Y$ is given in the following table.

$y$	$0$	$1$	$2$	$3$	$4$
$P (Y = y)$	$0.03$	$2 p$	$0.32$	$p$	$0.05$

Find the value of $p .$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 6 - Probability Distributions>EXERCISE 6B>Q 2

The probability distribution for the random variable $Y$ is given in the following table.

$y$	$0$	$1$	$2$	$3$	$4$
$P (Y = y)$	$0.03$	$2 p$	$0.32$	$p$	$0.05$

Calculate $E (Y)$ and the standard deviation of $Y .$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 6 - Probability Distributions>EXERCISE 6B>Q 3

The random variable

T

is such that

T \in \{1, 3, 6, 10\}

Given that the four possible values of

T

are equiprobable, find

E (T)

and Var

(T)

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 6 - Probability Distributions>EXERCISE 6B>Q 4

The following table shows the probability distribution for the random variable $V .$

$y$	$1$	$3$	$9$	$m$
$P (V = v)$	$0.4$	$0.28$	$0.14$	$0.18$

Given that $E (V) = 5.38,$ find the value of $m$ and calculate Var $(V) .$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 6 - Probability Distributions>EXERCISE 6B>Q 5

R

is a random variable such that

R \in \{10, 20, 70, 100\} .

Given that

P (R = r)

is proportional to

r,

show that

E (R) = 77

and find Var

(R)

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 6 - Probability Distributions>EXERCISE 6B>Q 6

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

$w$	$2$	$7$	$a$	$24$
$P (W = w)$	$0.3$	$0.3$	$0.1$	$0.3$

Given that $E (W) = a$ find $a$ and evaluate Var $(W)$

The discrete random variable R is such that R∈1, 3, 5, 7 Given that PR=r=kr+1r+2, find the value of the constant k.

Important Questions on Probability Distributions

Learn and Prepare for any exam you want !

The discrete random variable $R$ is such that $R \in \{1, 3, 5, 7\}$

Given that $P (R = r) = \frac{k (r + 1)}{r + 2},$ find the value of the constant $k .$