Selection of Coloured Balls – 01

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red ball

Example – 01:

  • An urn contains 9 red, 7 white, and 4 black balls.  All balls are identical. If one ball is drawn at random from the urn. Find the probability that
  • Solution:
9 R 7W 4 B

Total = 9 + 7 + 4 = 20

There are 20 balls in the urn

one ball out of 20 can be drawn by 20C1 ways

Hence n(S) = 20C1 = 20



  • a red ball

Let A be the event of getting a red ball

There are 9 red balls in the urn

1 red ball out of 9 red balls can be drawn by 9Cways

∴ n(A) = 9C1  = 9



By the definition P(A) = n(A)/n(S) =9/20

Therefore the probability of getting a red ball is 9/20.

  • a white ball

Let B be the event of getting a white ball

There are 7 white balls in the urn

1 white ball out of 7 white balls can be drawn by 7Cways



∴ n(B) = 7C1  = 7

By the definition P(B) = n(B)/n(S) =7/20

Therefore the probability of getting a white ball is 7/20.

  • a black ball

Let C be the event of getting a black ball

There are 4 black balls in the urn



1 black ball out of 4 black balls can be drawn by 4Cways

∴ n(C) = 4C1  = 4

By the definition P(C) = n(C)/n(S) = 4/20 = 1/5

Therefore the probability of getting a black ball is 1/5.

  • not a red ball

Let D be the event of getting not a red ball



There are 7 + 4 = 11 non-red balls

1 non-red ball out of 11 non-red balls can be drawn by 11Cways

∴ n(D) = 11C1  = 11

By the definition P(D) = n(D)/n(S) = 11/20



Therefore the probability of getting not red ball is 11/20

  • not a white ball

Let E be the event of getting not a white ball

There are 9 + 4 = 13  non-white balls

1 non-white ball out of 13 non-white balls can be drawn by 13Cways

∴ n(E) = 13C1  = 13

By the definition P(E) = n(E)/n(S) = 13/20



Therefore the probability of getting not white ball is 13/20

  • not a black ball

Let F be the event of getting not a black ball

There are 9 + 7 = 16 non-black balls

1 non-black ball out of 16 non-black balls can be drawn by 16Cways



∴ n(F) = 16C1  = 16

By the definition P(F) = n(F)/n(S) = 16/20 = 4/5

Therefore the probability of getting not a black ball is 4/5.

  • a red ball or a black ball

Let G be the event of getting not a red ball

There are 9 + 4 = 13 red or black balls

1 red or a  black ball out of 13 can be drawn by 13Cways



∴ n(G) = 13C1  = 13

By the definition P(G) = n(G)/n(S) = 13/20

Therefore the probability of getting a red ball or a black ball is 13/20.

  • a red ball or a white ball

Let H be the event of getting a red ball or a white ball

There are 9 + 7 = 16 red or white balls

1 red or a  white ball out of 16 can be drawn by 16Cways

∴ n(H) = 16C1  = 16

By the definition P(H) = n(H)/n(S) = 16/20 = 4/5

Therefore the probability of getting a red ball or a white ball is 4/5.

  • a black ball or a white ball

Let J be the event of getting a black ball or a white ball

There are 4 + 7 = 11 black or white balls

1 black or a  white ball out of 11 can be drawn by 11Cways

∴ n(J) = 11C1  = 11

By the definition P(J) = n(J)/n(S) = 11/20

Therefore the probability of getting a black ball or a white ball is 11/20.

Example – 02:

  • An urn contains 9 red, 7 white, and 4 black balls.  All balls are identical. Two balls are drawn at random from the urn. Find the probability that
  • Solution:
9 R 7W 4 B

Total = 9 + 7 + 4 = 20

There are 20 balls in the urn

Two balls out of 20 can be drawn by 20C2 ways

Hence n(S) = 20C2 = 10 x 19

  • both red balls

Let A be the event of getting both red balls

There are 9 red balls in the urn

2 red balls out of 9 red balls can be drawn by 9Cways

∴ n(A) = 9C2  = 9 x 4

By the definition P(A) = n(A)/n(S) = (9 x 4)/(10 x 19) = 18/95

Therefore the probability of getting both red balls is 18/95

  • no red ball

Let B be the event of getting no red ball

There are 7 + 4 = 11 non-red balls in the urn

2 red balls out of 11 non-red balls can be drawn by 11Cways

∴ n(B) = 11C2  = 11 x 5

By the definition P(B) = n(B)/n(S) = (11 x 5)/(10 x 19) = 11/38

Therefore the probability of getting no red ball is 11/76

  • atleast one red ball

Let C be the event of getting atleast one red ball

Hence C is the event of getting no red ball

There are 7 + 4 = 11 non-red balls in the urn

2 non-red balls out of 11 non-red balls can be drawn by 11Cways

∴ n(C) = 11C2  = 11 x 5

By the definition P(C) = n(C)/n(S) = (11 x 5)/(10 x 19) = 11/38

Now, P(C) = 1 – P(C) = 1 – 11/38 = 27/38

Therefore the probability of getting atleast one red ball is 27/38

  • exactly one red ball

Let D be the event of getting exactly one red ball i.e. one red and 1 non red ball

There are 9 red and 7 + 4 = 11 non-red balls in the urn

2 red balls out of 11 non-red balls can be drawn by 11Cways

∴ n(D) = 9C1  x  11C1  = 9  x 11

By the definition P(B) = n(B)/n(S) = (9 x 11)/(10 x 19) = 99/190

Therefore the probability of getting exactly one red ball is 99/190.

  • at most one red ball

Let E be the event of getting at most one red ball There are two possibilities

Case – 1: Getting no red ball and two non-red balls or

Case – 2: Getting 1 red ball and 1 non-red ball

There are 9 red and 7 + 4 = 11 non-red balls in the urn

∴ n(E) = 9C0  x  11C2 +  9C1  x  11C= 1 x 11 x 5 + 9 x 11 = 55 + 99 = 154

By the definition P(E) = n(E)/n(S) = 154/(10 x 19) = 77/95

Therefore the probability of getting at most one red ball is 77/95.

  • one is red and other is white

Let F be the event of getting one red and one white ball

There are 9 red and 7 white balls in the urn

∴ n(F) = 9C1  x  7C1 = 9 x 7

By the definition P(F) = n(F)/n(S) = (9 x 7)/(10 x 19) = 63/190

Therefore the probability of getting one red and other white ball is 63/190

  • one is red and other is black

Let G be the event of getting one red and one black ball

There are 9 red and 4 black balls in the urn

∴ n(G) = 9C1  x  4C1 = 9 x 4

By the definition P(G) = n(G)/n(S) = (9 x 4)/(10 x 19) = 18/95

Therefore the probability of getting one red and other black ball is 18/95

  • one is white and other is black

Let H be the event of getting one white and one black ball

There are 7 white and 4 black balls in the urn

∴ n(H) = 7C1  x  4C1 = 7 x 4

By the definition P(H) = n(H)/n(S) = (7 x 4)/(10 x 19) = 14/95

Therefore the probability of getting one white and other black ball is 14/95.

  • both are of same colour

Let J be the event of getting balls of same colour.

i.e. both are red or both are white or both are black

There are 9 red, 7 white, and 4 black balls in the urn

∴ n(J) = 9C2  +  7C2 +  4C2 = 9 x 4 + 7 x 3 + 2 x 3 = 36 +21 + 6 = 63

By the definition P(J) = n(J)/n(S) = 63/(10 x 19) = 63/190

Therefore the probability of getting both balls of same colour is 63/190

  • both are of not of the same colour

Let K be the event of getting balls not of the same colour.

Hence K is the event of getting the balls of same colour

i.e. both are red or both are white or both are black

There are 9 red, 7 white, and 4 black balls in the urn

∴ n(K) = 9C2  +  7C2 +  4C2 = 9 x 4 + 7 x 3 + 2 x 3 = 36 +21 + 6 = 63

By the definition P(K) = n(K)/n(S) = 63/(10 x 19) = 63/190

Now, P(K) = 1 – P(K) = 1 – 63/190 = 127/190

Therefore the probability of getting both balls of not of the same colour is 127/190.

  • both are red or both are black

Let L be the event of getting both red or both black balls.

There are 9 red and 4 black balls in the urn

∴ n(L) = 9C2 +  4C2 = 9 x 4 + 2 x 3 = 36 + 6 = 42

By the definition P(L) = n(L)/n(S) = 42/(10 x 19) = 21/95

Therefore the probability of getting both red or both black balls is 21/95.

Science > Mathematics > ProbabilityYou are Here
Physics Chemistry  Biology  Mathematics

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