Conditional probability in real world problems worksheets, solved worksheet problems or exercises with step-by-step work, formative assessment as online test, calculator and more learning resources to learn, practice, assess and master the basic math skills of **statistics**.

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1

Given that E and F are the two events such that P(E)= 0.6, P(F)= 0.3 and P(E∩F)= 0.2, find (i)P(E|F) and (ii)P(F|E)

2

If P(A)= 0.8, P(B)= 0.5 and P(B|A)= 0.4, find (i)P(A∩B) (ii)P(A|B) (iii)P(A∪B)

3

A fair die is rolled. Consider events E= {1, 3, 5}, F= {2, 3} and G ={2, 3, 4, 5} Find (i)P(E|F) (ii)P(E|G)

4

A family has two children. What is the probability that both the children are boys given that at least one of them is a boy ?

5

A die is thrown twice and the sum of the numbers appearing is observed to be 7. What is the conditional probability that the number 4 has appeared at least once?

6

Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

7

A die is thrown three times. Event A defined as 4 on the third throw and Event B defined as 6 on the first and 5 on the second throw. Find the probability of A given that B has already occurred.

8

Two coins are tossed once, where (i)E:tail appears on one coin, F:one coin shows head (ii)E:no tail appears, F:no head appears

9

A coin is tossed three times, where E:head on third toss , F:heads on first two tosses

10

Mother, father and son line up at random for a family picture. E:son on one end, F:father in middle

Answers Key

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1

Given that E and F are the two events such that P(E)= 0.6, P(F)= 0.3 and P(E∩F)= 0.2, find (i)P(E|F) and (ii)P(F|E)

(i) P(E|F) is 2/3

(ii) P(F|E) is 1/3

(ii) P(F|E) is 1/3

Solution

**(i) **Condition probability of E given F, P(E|F)= P(E∩F)/P(F)

**(ii) **Condition probability of F given E, P(F|E)= P(E∩F)/P(E)

__Result__

(i) P(E|F) is 2/3

(ii) P(F|E) is 1/3

Given P(E)= 0.6, P(F)= 0.3 and P(E∩F)= 0.2

P(E|F)= 0.2/0.3

P(E|F)= 2/3

Conditional probability, P(E|F)= 2/3

P(F|E)= 0.2/0.6

P(F|E)= 2/6

Conditional probability, P(F|E)= 1/3

(i) P(E|F) is 2/3

(ii) P(F|E) is 1/3

2

If P(A)= 0.8, P(B)= 0.5 and P(B|A)= 0.4, find (i)P(A∩B) (ii)P(A|B) (iii)P(A∪B)

(i) P(A∩B) is 0.32

(ii) P(A|B) is 0.64

(iii) P(A∪B) is 0.98

(ii) P(A|B) is 0.64

(iii) P(A∪B) is 0.98

Solution

**(i) **Condition probability of B given A, P(B|A)= P(A∩B)/P(A)

**(ii) **Condition probability of A given B, P(A|B)= P(A∩B)/P(B)

**(iii) **P(A∪B)= P(A)+P(B)- P(A∩B)

__Result__

(i) P(A∩B) is 0.32

(ii) P(A|B) is 0.64

(iii) P(A∪B) is 0.98

Given P(A)= 0.8, P(B)= 0.5 and P(B|A)= 0.4

∴ P(A∩B)= P(B|A)*P(A)

P(A∩B)= 0.4*0.8

P(A∩B)= 0.32

P(A|B)= 0.32/0.5

P(A|B)= 0.64

P(A∪B)= 0.8+0.5- 0.32

P(A∪B)= 0.98

(i) P(A∩B) is 0.32

(ii) P(A|B) is 0.64

(iii) P(A∪B) is 0.98

3

A fair die is rolled. Consider events E= {1, 3, 5}, F= {2, 3} and G ={2, 3, 4, 5} Find (i)P(E|F) (ii)P(E|G)

(i) P(E|F) is 0.5

(ii) P(E|G) is 0.5

(ii) P(E|G) is 0.5

Solution

**(i) **Condition probability of E given F, P(E|F)= P(E∩F)/P(F)

**(ii) **Condition probability of E given G, P(E|G)= P(E∩G)/P(G)

__Result__

(i) P(E|F) is 0.5

(ii) P(E|G) is 0.5

Sample space for a die, S is {1, 2, 3, 4, 5, 6}

Given E= {1, 3, 5}, F= {2, 3} and G= {2, 3, 4, 5}

∴ n(S)= 6, n(E)= 3, n(F)= 2, n(G)= 4

∴ P(E)= 3/6, P(F)= 2/6, P(G)= 4/6

∴ P(E)= 3/6, P(F)= 2/6, P(G)= 4/6

Set of elements in set E that are also present in set F, (E∩F)= {3}

Set of elements in set E that are also present in set G, (E∩G)= {3, 5}

Set of elements in set E that are also present in set G, (E∩G)= {3, 5}

∴ n(E∩F)= 1, n(E∩G)= 2

∴ P(E∩F)= 1/6, P(E∩G)= 2/6

P(E|F)= (1/6)/(2/6)

Conditional probability, P(E|F)= 0.5

P(E|G)= (2/6)/(4/6)

Conditional probability, P(E|G)= 0.5

(i) P(E|F) is 0.5

(ii) P(E|G) is 0.5

4

A family has two children. What is the probability that both the children are boys given that at least one of them is a boy ?

Probability that both the children are boys given that at least one of them is a boy is 1/3

Solution

Let b stand for boy and g for girl.

Let E be the event that both the children are boys.

Let F be the event that at least one of the child is a boy.

Sample space of the experiment, S= {(b, b), (g, b), (b, g), (g, g)}

Let F be the event that at least one of the child is a boy.

Sample space of the experiment, S= {(b, b), (g, b), (b, g), (g, g)}

∴ E= {(b,b)} and F= {(b,b), (g,b), (b,g)}

∴ n(S)= 4, n(E)= 1, n(F)= 3, P(E)= 1/4, P(F)= 3/4

Set of elements in set E that are also present in set F, E∩F= {(b,b)}

∴ n(S)= 4, n(E)= 1, n(F)= 3, P(E)= 1/4, P(F)= 3/4

Set of elements in set E that are also present in set F, E∩F= {(b,b)}

∴ n(E∩F) = 1, P(E∩F) = 1/4

Condition probability of E given F, P(E|F)= P(E∩F)/P(F)

P(E|F)= (1/4)/(3/4)

P(E|F)= (1/4)/(3/4)

∴ Probability that both the children are boys given that at least one of them is a boy= 1/3

5

A die is thrown twice and the sum of the numbers appearing is observed to be 7. What is the conditional probability that the number 4 has appeared at least once?

Conditional probability that the number 4 has appeared at least once is 0.3333

Solution

Sample space for a die thrown twice, n(S)= 36.

Let E be the event that number 4 appears at least once.

Let F be the event that the sum of the numbers appearing is 7.

Let F be the event that the sum of the numbers appearing is 7.

E= {(1,4)(2,4)(3,4)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,4)(6,4)}

F= {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)}

∴ n(E)= 11 , n(F)= 6

∴ P(E)= 11/36, P(F)= 6/36

∴ P(E)= 11/36, P(F)= 6/36

Set of elements in set E that are also present in set F, E∩F= {(3,4)(4,3)}

∴ n(E∩F)= 2, P(E∩F)= 2/36

∴ n(E∩F)= 2, P(E∩F)= 2/36

Condition probability of E given F, P(E|F)= P(E∩F)/P(F)

P(E|F)= (2/36)/(6/36)

P(E|F)= (2/36)/(6/36)

P(E|F)= 2/6

∴ Conditional probability that number 4 has appeared at least once= 0.3333

6

Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Probability that drawn card is an even number is 4/7

Solution

The sample space of the experiment, S= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Let A be the event the number on the card drawn is even.

Let B be the event the number on the card drawn is greater than 3.

Let A be the event the number on the card drawn is even.

Let B be the event the number on the card drawn is greater than 3.

A= {2, 4, 6, 8, 10}

B= {4, 5, 6, 7, 8, 9, 10}

∴ n(S)= 10 , n(A)= 5, n(B)= 7

∴ P(A)= 5/10, P(B)= 7/10

∴ P(A)= 5/10, P(B)= 7/10

Set of elements in set A that are also present in set B, A∩B= { 4, 6, 8, 10}

∴ n(A∩B)= 4, P(A∩B)= 4/10

∴ n(A∩B)= 4, P(A∩B)= 4/10

Condition probability of A given B, P(A|B)= P(A∩B)/P(B)

P(A|B)= (4/10)/(7/10)

P(A|B)= (4/10)/(7/10)

P(A|B)= 4/7

∴ Probability that drawn card is an even number= 4/7

7

A die is thrown three times. Event A defined as 4 on the third throw and Event B defined as 6 on the first and 5 on the second throw. Find the probability of A given that B has already occurred.

∴ Probability of A given B is 1/6

Solution

Total sample space of a die thrown three times, n(S)= 216.

Event A defined as 4 on the third throw.

Event B defined as 6 on the first and 5 on the second throw.

Event A defined as 4 on the third throw.

Event B defined as 6 on the first and 5 on the second throw.

Possible outcomes of 4 on the third throw, event A is given below,

(1,1,4) , (1,2,4) , (1,3,4) , (1,4,4) , (1,5,4) , (1,6,4)

(2,1,4) , (2,2,4) , (2,3,4) , (2,4,4) , (2,5,4) , (2,6,4)

(3,1,4) , (3,2,4) , (3,3,4) , (3,4,4) , (3,5,4) , (3,6,4)

(4,1,4) , (4,2,4) , (4,3,4) , (4,4,4) , (4,5,4) , (4,6,4)

(5,1,4) , (5,2,4) , (5,3,4) , (5,4,4) , (5,5,4) , (5,6,4)

(6,1,4) , (6,2,4) , (6,3,4) , (6,4,4) , (6,5,4) , (6,6,4)

B = {(6,5,1) , (6,5,2) , (6,5,3) , (6,5,4) , (6,5,5) , (6,5,6) }

∴ n(S)= 216 , n(A)= 36, n(B)= 6

∴ P(A)= 36/216, P(B)= 6/216

∴ P(A)= 36/216, P(B)= 6/216

Set of elements in set A that are also present in set B, A∩B= { (6,5,4) }

∴ n(A∩B)= 1, P(A∩B)= 1/216

∴ n(A∩B)= 1, P(A∩B)= 1/216

Condition probability of A given B, P(A|B)= P(A∩B)/P(B)

P(A|B)= (1/216)/(6/216)

P(A|B)= (1/216)/(6/216)

P(A|B)= 1/6

∴ Probability of A given B= 1/6

8

Two coins are tossed once, where (i)E:tail appears on one coin, F:one coin shows head (ii)E:no tail appears, F:no head appears

(i) P(E|F) is 1

(ii) P(E|F) is 0

(ii) P(E|F) is 0

Solution

**(i) ** E : tail appears on one coin, F : one coin shows head

**(ii) ** E : no tail appears, F : no head appears

__Result__

(i) P(E|F)= 1

(ii) P(E|F) = 0

The sample space of the experiment, S= {HH,HT,TH,TT}

E= {HT,TH}, F= {HT,TH}

∴ n(S)= 4 , n(E)= 2, n(F)= 2

∴ P(A)= 2/4, P(B)= 2/4

∴ P(A)= 2/4, P(B)= 2/4

Set of elements in set E that are also present in set F, E∩F= {HT,TH}

∴ n(E∩F)= 2, P(A∩B)= 2/4

∴ n(E∩F)= 2, P(A∩B)= 2/4

Condition probability of E given F, P(E|F)= P(E∩F)/P(F)

P(E|F)= (2/4)/(2/4)

P(E|F)= 1

P(E|F)= (2/4)/(2/4)

P(E|F)= 1

E= {HH}, F= {TT}

∴ n(S)= 4 , n(E)= 1, n(F)= 1

∴ P(A)= 1/4, P(B)= 1/4

∴ P(A)= 1/4, P(B)= 1/4

Set of elements in set E that are also present in set F, E∩F= 0

∴ n(E∩F)= 0, P(A∩B)= 0/4=0

∴ n(E∩F)= 0, P(A∩B)= 0/4=0

Condition probability of E given F, P(E|F)= P(E∩F)/P(F)

P(E|F)= (0)/(1/4)

P(E|F)= 0

P(E|F)= (0)/(1/4)

P(E|F)= 0

(i) P(E|F)= 1

(ii) P(E|F) = 0

9

A coin is tossed three times, where E:head on third toss , F:heads on first two tosses

P(E|F) is 1/2

Solution

The sample space, S= {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}

E : head on third toss.

F : heads on first two tosses.

F : heads on first two tosses.

E= {HHH,HTH,THH,TTH}, F= {HHH,HHT}

∴ n(S)= 8 , n(E)= 4, n(F)= 2

∴ P(A)= 4/8, P(B)= 2/8

∴ P(A)= 4/8, P(B)= 2/8

Set of elements in set E that are also present in set F, E∩F= {HHH}

∴ n(E∩F) = 1, P(A∩B)= 1/8

∴ n(E∩F) = 1, P(A∩B)= 1/8

Condition probability of E given F, P(E|F)= P(E∩F)/P(F)

P(E|F)= (1/8)/(2/8)

P(E|F)= 1/2

P(E|F)= (1/8)/(2/8)

P(E|F)= 1/2

∴ P(E|F)= 1/2

10

Mother, father and son line up at random for a family picture. E:son on one end, F:father in middle

P(E|F) is 1

Solution

The sample space, S= {MFS, SFM, FSM, MSF, SMF, FMS}

Let M denote mother, F denote father and S denote son.

∴E= {MFS, SFM, SMF, FMS} and F= {MFS, SFM}

∴ n(S)= 6, n(E)= 4, n(F)= 2, P(E)= 4/6, P(F)= 2/6

Set of elements in set E that are also present in set F, E∩F= {MFS, SFM}

∴ n(S)= 6, n(E)= 4, n(F)= 2, P(E)= 4/6, P(F)= 2/6

Set of elements in set E that are also present in set F, E∩F= {MFS, SFM}

∴ n(E∩F)= 2, P(E∩F)= 2/6

Condition probability of E given F, P(E|F)= P(E∩F)/P(F)

P(E|F)= (2/6)/(2/6)

P(E|F)= (2/6)/(2/6)

∴ P(E|F)= 1

Online Test

Start my Challenge

A coin is tossed three times, where E : head on third toss , F : heads on first two tosses

If P(A) = 0.8, P (B) = 0.8 and P(B|A) = 0.3, find (i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)