Answer :

There are 4 white, 7 black and 5 red balls in the bag, so the number of all favorable outcomes in the sample space is

n(S) = 4+7+5=16

Let A be the event of getting a white ball in the first draw. Hence the probability becomes

(as there are 4 white balls out of 16 balls)

Let B represents the event of getting a black ball in the second draw. Hence the probability becomes

(as there are 7 black balls out of 15 balls as balls are not replaced back in the bag)

Let C represents the event of getting a red ball in the third draw. Hence the probability becomes

(as there are 5 red balls out of 14 balls as balls are not replaced back in the bag)

Then the probability that the balls drawn are white, black and red respectively without replacement

Hence the required probability is

Rate this question :

A bag contains 6 RD Sharma - Volume 2

A man is known toMathematics - Board Papers

A bag contaRD Sharma - Volume 2

Three persons A, Mathematics - Board Papers

Bag I contains 3 Mathematics - Board Papers

A speaks trRD Sharma - Volume 2

Two cards aRD Sharma - Volume 2

A bag contaRD Sharma - Volume 2

Kamal and MRD Sharma - Volume 2

Two balls aRD Sharma - Volume 2