# In her pocket, Kira has 2 red marbles, 2 green marbles and 2 blue marbles that are all the same size.In her pocket, Kira has 2 red marbles, 2 green marbles and 2 blue marbles that are all the same...

In her pocket, Kira has 2 red marbles, 2 green marbles and 2 blue marbles that are all the same size.

In her pocket, Kira has 2 red marbles, 2 green marbles and 2 blue marbles that are all the same size. If Kira picks one marble out of her pocket without looking, what is the probabliliy that it will be either red or green?

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

red marbles = 2

green marbles = 2

blue marbles = 2

Total marbles = 6

We need tp calculate the probability of getting a red OR green

We know that:

P(red) = 2/6 = 1/3

P(green) = 2/6= 1/3

P(blue) = 2/6 = 1/3

P(red or green) = P(red ) + P(green)

= 1/3  +  1/3 = 2/3

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

As given in the question, Kira has 2 red marbles, 2 green marbles and 2 blue marbles. Therefore she has a total of 2+2+2=6 marbles in her pocket and 2 of those are blue which is not red or green.

Therefore the probability that a marble picked out of her pocket is not red or green is 2/6= 1/3.

As the (probability of an event not happening) is 1- (probability of the event happening), the probability that a marbled picked at random being either red or green is 1-(1/3)= (2/3).

The probability that a marbled picked at random is either red or green is (2/3).

neela | High School Teacher | (Level 3) Valedictorian

Posted on

It is clear that any of the 6 balls are equally likey ia draw.

The probabilty of an event is = number chances favouarable  to the event/ total number of ways (favourable  and unfavourable to the event)

The probabilty drawing green P(G) = 2/6.

Probability of drawing red : P(R) = 2/6

So P(G or R) = P(G)+P(R) - P(G & B) = 1/6+1/6+0 = 4/6 = 2/3.

P(G & R) = 0 , as both  colour balls can not  happen.