Question

A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken
out of the box at random. What is the probability that the marble taken out will be
(i) red? (ii) white? (iii) not green?​

Answer 1

Answer:

sample space (S) = (5 red + 8 white + 4 green)

n(S) = 17

Evevt A : Red marbles

A : ( 5 red marbles )

n(A) = 5

p(A) = n(A) / n(S)

= 5 / 17

p(A) = 5 / 17

Evevt B : White marbles

B : ( 8 marbles )

n(B) = 8

p(B) = n(B) / n(S)

= 8 / 17

p(B) = 8 / 17

Event C : Not green marbles

C : ( 5 red + 8 white )

n(C) = 13

p(C) = n(C) / n(S)

= 13 / 17

p(C) = 13 / 17

I hope this will help you....

Answer 2

ANSWER

Number of red marbles in the box = 5 .

Number of white marbles in the box = 8 .

Number of green marbles in the box = 4 .

$\boxed{\sf Probability \:of \:an \:Event = \dfrac{Number \: of \: Favorable \: Outcomes}{Total \:Number \:of \:Outcomes} }$

Total Number of Outcomes = Total Number of marbles

$\implies$ 5 + 8 + 4  

∴ Total Number of Outcomes = 17.

If one marble is drawn out then , Probability of getting

(i) red marble

$\sf Probability (red \: marble ) = \dfrac{Number \: of \: red \: marbles}{Total \: Number \: of \: marbles}$

Number of red marble = 5.

Total Number of Marbles = 17.

$\red\bigstar \sf Probability (red \: marble ) = \dfrac{5}{17}$

(ii) white marble

$\sf Probability (white \: marble ) = \dfrac{Number \: of \: white \: marbles}{Total \: Number \: of \: marbles}$

Number of white marble = 8

Total Number of Marbles = 17

$\red\bigstar \sf Probability (white \: marble ) = \dfrac{8}{17}$

(iii) not green

Probability(not green marble) = Probability (red marble) + Probability (white marble)

$\implies \sf Probability(not \: green \: marble) = \dfrac{5}{17} + \dfrac{8}{17}$

$\red\bigstar \sf Probability(not \: green \: marble) = \dfrac{13}{17}$