**Probability Word Problems with Three Events :**

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

**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**

**consecutive two heads**

**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)+3****P(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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