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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?
A bag has 100 balls out of which there is one green ball. 50 balls are randomly selected. What is probability that the green ball is selected.
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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