The probability distribution for the random variable W is given in the following table. w 3 6 9 12 15 P W = w 2 k k 2 k 2 4 5 - 3 k 13 50 Form an equation using k , then solve it.

Given that, The probability distribution for the random variable W is given in the following table. w 3 6 9 12 15 P W = w 2 k k 2 k 2 4 5 - 3 k 13 50 Given that, ∑ i = 1 n P i = 1 ∴ P ( W = 3 ) + P ( W = 6 ) + D ( W = 9 ) + P ( W = 12 ) + P ( W = 15 ) = 1 ⇒ 2 k + k 2 + k 2 + 4 5 - 3 k + 13 50 = 1 ⇒ k 2 - k 2 + 53 50 = 1 ⇒ 50 k 2 - 25 k + 3 = 0 ⇒ 50 k 2 - 10 k - 15 k + 3 = 0 ⇒ 10 k ( 5 k - 1 ) - 3 ( 5 k - 1 ) = 0 ∴ ( 10 k - 3 ) ( 5 k - 1 ) = 0 ∴ k = 3 10 ond k = 1 5 ∴ k = 0 . 3 and k = 0 . 2 We know that, 0 ⩽ p ≤ 1 ∴ k = 0 . 3 is not possible ∴ k = 0 . 2 is the only valid solution ⇒ P ( 6 ≤ W < 10 ) = P ( W = 6 ) + p ( W = 9 ) ⇒ k 2 + k 2 = ( 0 . 2 ) 2 + 0 . 2 2 = 0 . 04 + 0 . 1 ⇒ P ( 6 ≤ W < 10 ) = 0 . 14

The probability distribution for the random variable W is given in the following table. w 3 6 9 12 15 P W = w 2 k k 2 k 2 4 5 - 3 k 13 50 Explain why only one of your solutions is valid.

Given that, The probability distribution for the random variable W is given in the following table. w 3 6 9 12 15 P W = w 2 k k 2 k 2 4 5 - 3 k 13 50 Given that, ∑ i = 1 n P i = 1 ∴ P ( W = 3 ) + P ( W = 6 ) + D ( W = 9 ) + P ( W = 12 ) + P ( W = 15 ) = 1 ⇒ 2 k + k 2 + k 2 + 4 5 - 3 k + 13 50 = 1 ⇒ k 2 - k 2 + 53 50 = 1 ⇒ 50 k 2 - 25 k + 3 = 0 ⇒ 50 k 2 - 10 k - 15 k + 3 = 0 ⇒ 10 k ( 5 k - 1 ) - 3 ( 5 k - 1 ) = 0 ∴ ( 10 k - 3 ) ( 5 k - 1 ) = 0 ∴ k = 3 10 ond k = 1 5 ∴ k = 0 . 3 and k = 0 . 2 If k = 0 . 3 , P ( W = 12 ) = 4 5 - 3 k ⇒ P W = 12 = 4 5 - 3 × 0 . 3 ⇒ P W = 12 = 0 . 8 - 0 . 9 = - 0 . 1 We know that, 0 ⩽ p ≤ 1 ∴ k = 0 . 3 is not possible ∴ k = 0 . 2 is the only valid solution

EASY

AS and A Level

IMPORTANT

Earn 100

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

$x$ $2$ $3$ $4$ $5$

$P (X = x)$ $p$ $2 p$ $\frac{1}{2} p$ $3 p$

Find the value of $p$ and work out $P (2 < X < 5)$

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 3

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

$w$	$3$	$6$	$9$	$12$	$15$
$P (W = w)$	$2 k$	$k^{2}$	$\frac{k}{2}$	$\frac{4}{5} - 3 k$	$\frac{13}{50}$

Form an equation using $k,$ then solve it.

EASY

AS and A Level

IMPORTANT

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

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

$w$	$3$	$6$	$9$	$12$	$15$
$P (W = w)$	$2 k$	$k^{2}$	$\frac{k}{2}$	$\frac{4}{5} - 3 k$	$\frac{13}{50}$

Explain why only one of your solutions is valid.

EASY

AS and A Level

IMPORTANT

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

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

$w$	$3$	$6$	$9$	$12$	$15$
$P (W = w)$	$2 k$	$k^{2}$	$\frac{k}{2}$	$\frac{4}{5} - 3 k$	$\frac{13}{50}$

Find $P (6 ⩽ W < 10)$

EASY

AS and A Level

IMPORTANT

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

The probability that a boy succeeds with each basketball shot is

\frac{7}{9} .

He takes two shots and the discrete random variable $S$ represents the number of successful shots. Show that

P (S = 0) = \frac{4}{81}

and draw up the probability distribution table for

S .

EASY

AS and A Level

IMPORTANT

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

At a garden centre, there is a display of roses: $25$ are red, $20$ are white, $15$ are pink and $5$ are orange. Three roses are chosen at random. Show that the probability of selecting three red roses is approximately $0.0527 .$

EASY

AS and A Level

IMPORTANT

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

At a garden centre, there is a display of roses: $25$ are red, $20$ are white, $15$ are pink and $5$ are orange. Three roses are chosen at random.

Draw up the probability distribution table for the number of red roses selected.

EASY

AS and A Level

IMPORTANT

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

At a garden centre, there is a display of roses: $25$ are red, $20$ are white, $15$ are pink and $5$ are orange. Three roses are chosen at random.

Find the probability that at least one red rose is selected.

EASY

AS and A Level

IMPORTANT

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

Three vehicles from a company's six trucks, five vans, three cars and one motorbike are randomly selected and tested for roadworthiness.

Show that the probability of selecting three vans is $\frac{2}{91} .$

The probability distribution for the random variable X is given in the following table. x 2 3 4 5 PX=x p 2p 12p 3p Find the value of p and work out P2<X<5

Important Questions on Probability Distributions

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

$x$ $2$ $3$ $4$ $5$

$P (X = x)$ $p$ $2 p$ $\frac{1}{2} p$ $3 p$

Find the value of $p$ and work out $P (2 < X < 5)$