Question

From a bag containing 5white, 7red and 4 black balls a man draws (a) 3 balls and (b) 4 balls at random. Find the probability that among the balls drawn there is at least one ball of each colour.

Answer 1

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such that

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)=

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)= 16

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)= 1611

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)= 1611

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)= 1611

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)= 1611 Now, required probability = P(A∩B∩C∩D)

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)= 1611 Now, required probability = P(A∩B∩C∩D)=P(A)P(B)P(C)P(D)=(

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)= 1611 Now, required probability = P(A∩B∩C∩D)=P(A)P(B)P(C)P(D)=( 16

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)= 1611 Now, required probability = P(A∩B∩C∩D)=P(A)P(B)P(C)P(D)=( 1611

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)= 1611 Now, required probability = P(A∩B∩C∩D)=P(A)P(B)P(C)P(D)=( 1611

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)= 1611 Now, required probability = P(A∩B∩C∩D)=P(A)P(B)P(C)P(D)=( 1611 )

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such thatP(A)=P(B)=P(C)=P(D)There are 16 balls out of which 11 are not white. Therefore,P(A)= 1611 Now, required probability = P(A∩B∩C∩D)=P(A)P(B)P(C)P(D)=( 1611 ) 4..

hope this will help u

Answer 2

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such that

P(A)=P(B)=P(C)=P(D)

There are 16 balls out of which 11 are not white. Therefore,

P(A)=  11/16

Now, required probability = P(A∩B∩C∩D)

=P(A)P(B)P(C)P(D)=(11/16)^4