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#### 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 ^{20}C_{1}_{ }ways

Hence n(S) = ^{20}C_{1} = 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 ^{9}C_{1 }ways

∴ n(A) = ^{9}C_{1} = 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 ^{7}C_{1 }ways

∴ n(B) = ^{7}C_{1} = 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 ^{4}C_{1 }ways

∴ n(C) = ^{4}C_{1} = 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 ^{11}C_{1 }ways

∴ n(D) = ^{11}C_{1} = 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 ^{13}C_{1 }ways

∴ n(E) = ^{13}C_{1} = 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 ^{16}C_{1 }ways

∴ n(F) = ^{16}C_{1} = 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 ^{13}C_{1 }ways

∴ n(G) = ^{13}C_{1} = 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 ^{16}C_{1 }ways

∴ n(H) = ^{16}C_{1} = 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 ^{11}C_{1 }ways

∴ n(J) = ^{11}C_{1} = 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 ^{20}C_{2}_{ }ways

Hence n(S) = ^{20}C_{2} = 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 ^{9}C_{2 }ways

∴ n(A) = ^{9}C_{2} = 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 ^{11}C_{2 }ways

∴ n(B) = ^{11}C_{2} = 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 ^{11}C_{2 }ways

∴ n(C) = ^{11}C_{2} = 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 ^{11}C_{2 }ways

∴ n(D) = ^{9}C_{1} x ^{11}C_{1} = 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) = ^{9}C_{0} x ^{11}C_{2} + ^{9}C_{1} x ^{11}C_{1 }= 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) = ^{9}C_{1} x ^{7}C_{1} = 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) = ^{9}C_{1} x ^{4}C_{1} = 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) = ^{7}C_{1} x ^{4}C_{1} = 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) = ^{9}C_{2} + ^{7}C_{2} + ^{4}C_{2} = 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) = ^{9}C_{2} + ^{7}C_{2} + ^{4}C_{2} = 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) = ^{9}C_{2} + ^{4}C_{2} = 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 > Probability > You are Here |

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