# Conditional Probability in Real World Problems: Worksheets, Online Test and Calculators

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.

## Conditional Probability in Real World Problems

Name
Date

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

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

(i) Condition probability of E given F, P(E|F)= P(E∩F)/P(F)
P(E|F)= 0.2/0.3
P(E|F)= 2/3
Conditional probability, P(E|F)= 2/3

(ii) Condition probability of F given E, P(F|E)= P(E∩F)/P(E)
P(F|E)= 0.2/0.6
P(F|E)= 2/6
Conditional probability, P(F|E)= 1/3

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

(i) Condition probability of B given A, P(B|A)= P(A∩B)/P(A)
∴ P(A∩B)= P(B|A)*P(A)
P(A∩B)= 0.4*0.8
P(A∩B)= 0.32

(ii) Condition probability of A given B, P(A|B)= P(A∩B)/P(B)
P(A|B)= 0.32/0.5
P(A|B)= 0.64

(iii) P(A∪B)= P(A)+P(B)- P(A∩B)
P(A∪B)= 0.8+0.5- 0.32
P(A∪B)= 0.98

Result
(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
Solution
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

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}
∴ n(E∩F)= 1, n(E∩G)= 2
∴ P(E∩F)= 1/6, P(E∩G)= 2/6

(i) Condition probability of E given F, P(E|F)= P(E∩F)/P(F)
P(E|F)= (1/6)/(2/6)
Conditional probability, P(E|F)= 0.5

(ii) Condition probability of E given G, P(E|G)= P(E∩G)/P(G)
P(E|G)= (2/6)/(4/6)
Conditional probability, P(E|G)= 0.5

Result
(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)}

∴ 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(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)

∴ 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.

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

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

Condition probability of E given F, P(E|F)= P(E∩F)/P(F)
P(E|F)= (2/36)/(6/36)
P(E|F)= 2/6

∴ 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.

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

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

Condition probability of A given B, P(A|B)= P(A∩B)/P(B)
P(A|B)= (4/10)/(7/10)
P(A|B)= 4/7

∴ 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.

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

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

Condition probability of A given B, P(A|B)= P(A∩B)/P(B)
P(A|B)= (1/216)/(6/216)
P(A|B)= 1/6

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

(i) E : tail appears on one coin, F : one coin shows head
E= {HT,TH}, F= {HT,TH}
∴ n(S)= 4 , n(E)= 2, n(F)= 2
∴ 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

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

(ii) E : no tail appears, F : no head appears
E= {HH}, F= {TT}
∴ n(S)= 4 , n(E)= 1, n(F)= 1
∴ 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

Condition probability of E given F, P(E|F)= P(E∩F)/P(F)
P(E|F)= (0)/(1/4)
P(E|F)= 0

Result
(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.

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

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

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)= 1/2
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(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)= 1

## Worksheet: Conditional Probability in Real World Problems

Conditional probability in real world problems worksheet is the largest collection of practice problems and solved exercises which can be served as homework, classwork or assignment problems for the students to learn, practice, assess, iterate and master the skills of how to solve such statistics problems in basic mathematics.

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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
A die is thrown twice and the sum of the numbers appearing is observed to be 12. What is the conditional probability that the number 6 has appeared at least once?
Given that E and F are the two events such that P(E) = 0.9, P(F) = 0.8 and P(E ∩ F) = 0.6, find (i) P(E|F) and (ii) P(F|E)
Mother, father and son line up at random for a family picture. E : son on one end, F : father in middle
A die is thrown three times. Event A defined as 6 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.
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)
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 ?
Ten cards numbered 1 to 10 are placed in a box, 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?
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)

## Online Test: Conditional Probability in Real World Problems

This online test to find the conditional probability in real world problems is a formative assessment which can be used as homework, classwork or assignment problems to assess and improve the learner's math skills on statistics.

### Conditional Probability Calculator

Total Events n(S) :
Success Events n(A) :
Success Events n(B) :
 Total Events n(S) 52 Success Events n(A) 4 Success Events n(B) 3 P(A) 0.08 P(A') 0.92 P(B) 0.06 P(B') 0.94 P(A ∩ B) 0.0048 P(A U B) 0.1352 P(A | B) 0.08 P(B | A) 0.06
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Conditional probability calculator is a key tool for users to generate step-by-step work to quickly understand the calculation, solve the homework problems, solve the classwork problems, prepare the answer key document for summative and formative assessments or immediately verify the entire calculation for the user supplied input values.