# PROBABILITY WORD PROBLEMS WITH THREE EVENTS

Probability Word Problems with Three Events :

Here we are going to see, some practice problems on probability with three events.

## Probability Word Problems with Three Events - Questions

Question 1 :

A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or consecutive two heads.

Solution :

Sample space =  {HHH. HHT, HTH, HTT, THH, THT, TTH, TTT}

Let A, B and C be the events of getting two heads, atleast one tail and consecutive two heads respectively.

A  =  { HHT, HTH, THH}

n(A)  =  1

P(A)  =  n(A) / n(S)

P(A)  =  1/8

atleast one tail  =  0 tail, 1 tail

B  =  {HHT, HTH, THH}

n(B)  =  3

P(B)  =  n(B) / n(S)

P(B)  =  3/8

C = {HHH. HHT, THH}

n(C)  =  3

P(C)  =  n(C) / n(S)

P(C)  =  3/8

A n B  =  {HHT, HTH, THH}

n(AnB)  =  3

P(A n B)  =  3/8

B n C  =  {HHT}

n(B n C)  =  1

P(B n C)  =  1/8

C n A  =  {HHT, THH}

n(C n A)  =  2

P(C n A)  =  2/8

A n B n C  =  {HHT, THH}

n(A n B n C)  =  2

P(A n B n C)  =  2/8

P(AuBuC) = P(A)+P(B)+P(C)-P(AnB)-P(Bnc)-P(CnA)+P(AnBnC)

=  (1/8) + (3/8) + (3/8) - (3/8) - (1/8) - (2/8) + (2/8)

=  (1 + 3 + 3 - 3 - 1)/8

=  3/8

Question 2 :

If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if P(A n B) = 1/6, P(BnC) = 1/4, P(AnC) = 1/8, P(A U B U C) = 9/10, P(A n B n C) = 1/15, then find P(A), P(B) and P(C) ?

Solution :

P(A n B) = 1/6

P(B n C) = 1/4

P(A n C) = 1/8

P(A U B U C) = 9/10

P(A n B n C) = 1/15

P(B) = 2 P(A)

P(C) = 3 P(A)

P(AuBuC) = P(A)+P(B)+P(C)-P(AnB)-P(Bnc)-P(CnA)+P(AnBnC)

(9/10)  =  P(A)+2P(A)+3P(A)-(1/6) - (1/4) - (1/8) + (1/15)

6P(A)  =  (9/10)+(1/6)+(1/4)+(1/8)-(1/15)

6P(A)  =  (108 + 20 + 30 + 15 - 8)/120

6P(A)  =  165/120

P(A)  =  165/120(6)

P(A)  =   11/48

P(B)  =  2P(A) =  22/48  =  11/24

P(C)  =  3P(A)  =  3(11/48)  =  11/16

Let us look into the next example on "Probability Word Problems with Three Events".

Question 3 :

In a class of 35, students are numbered from 1 to 35. The ratio of boys to girls is 4:3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with prime roll number or a girl with composite roll number or an even roll number.

Solution :

Total numbers  =  1 to 35

n(S)  =  35

4x + 3x  =  35

7x  =  35

x  =  5

Number of boys  =  4x  =  4(5)  =  20

Number of girls  =  3x  =  3(5)  =  15

Let A, B and C be the  events of getting a boy with prime roll number or a girl with composite roll number or an even roll number.

Boys are numbered  = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

A boy with prime roll number = A = {2, 3, 5, 7, 11, 13, 17, 19}

n(A)  =  8

P(A)  =  8/35

Girls are numbered  =  {21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35}

A girl with composite roll number = B = {21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35}

n(B)  =  12

P(B)  =  12/35

An even roll number = C = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34}

n(C)  =  17

P(C)  =  17/35

A and B are mutually exclusive events.

P(A n B)  =  0

(B n C)  =  {22, 24, 26, 28, 30, 32, 34}

n(B n C)  =  7

P(B n C)  =  7/35

(C n A)  =  {2}

n(C n A)  =  1

P(C n A)  =  1/35

A n B n C  =  0

P(AuBuC) = P(A)+P(B)+P(C)-P(AnB)-P(Bnc)-P(CnA)+P(AnBnC)

P(AuBuC) = (8/35)+(12/35)+(17/35)-0-(7/35)-(1/35)+0

=  (8 + 12 + 17 - 7 - 1)/35

P(AuBuC)  =  29/35 After having gone through the stuff given above, we hope that the students would have understood, "Probability Word Problems with Three Events".

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