# A bag contains 5

There are 5 white, 7 red and 3 black balls in the bag, so the number of all favorable outcomes in the sample space is

n(S) = 5+7+3=15

Let A be the event of getting a red ball in the first draw. Hence the probability becomes (as there are 7 red balls out of 15 balls)

Let B represents the event of getting a red ball in the second draw. Hence the probability becomes (as there are 7 red balls and one red ball is already drawn in first draw so now there are total of 6 red balls)

Let C represents the event of getting a red ball in the third draw. Hence the probability becomes (as there are 7 red balls and two red balls are already drawn in first and second draw so now there are total of 5 red balls)

Then the probability of none being red balls without replacement     Hence the required probability is Rate this question :

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