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In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
total number of balls = (8 + 7 + 6) = 21.
Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue.
So n(E) = 7
P(E) = n(E)/n(S) =7/21
or P(E) = 1/3
7/(8+7+6) = 1/3
A bag contains 5 white, 4 red and 3 black balls. A ball is drawn at random from the bag, what is the probability that it is not black?
3 red, 2 white and 3 green balls are available in a bag. 2 balls are drawn one by one from the bag.
What will be the probability of one ball is white & other is green?
There are two bags. First contain 10 white balls and other contain 8 red balls. One ball from any bag shifted to other bag at random.
What will be the probability the second ball taken after first event would be red ?
First event is to drawn first ball.
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