A beg contains 4 white , 5 red and 6 green ball . Three balls are drawn at random . What is the chance that a white , a red and a green ball is drawn ?

2 Answers
Mar 27, 2018

#4/91#

Explanation:

Let's first see how many balls there are in total.

#4+5+6=15# total balls.

So, for the first draw, you want to draw a white ball. #4/15# of the balls are white, so this is the probability of drawing a white ball.

You want the next ball to be red. Since one has been drawn, only #14# remain, #5# of which are red. So the probability of drawing a red ball is #5/14#.

Finally, you want a green ball. Since two have been drawn, only #13# remain, 6 of which are green. So the probability of drawing a green ball is #6/13.#

The total probability is #4/15*5/14*6/13=120/2730=4/91#

Mar 27, 2018

Probability of taking one White, one Red and one Green ball#=24/91#

Explanation:

Give -
White Balls#=4#
Red Balls#=5#
Green Balls #=6#
Total #=15#

Probability of taking one White, one Red and one Green ball

#=(""^4C_1xx""^5C_1 xx ""^6C_1)/(""^15C_3)#

#=(4 xx 5 xx 6)/455=120/455=24/91#