Probability and Statistics Practice Problems

This page contains practice problems and solutions for probability and statistics covering topics such as conditional probability, Bayes' theorem, possibility space diagrams, and more. Each problem is explained step-by-step to help students improve their skills and knowledge.

Question 1
Using the rule of addition
P ( A ∪ B ¿=P ( A ) + P ( B ) −P( A ∩ B)
Where
P ( A ) =1−P ( A )
P ( A )=1− 1
4
P ( A )= 3
4
P ( B )= 1
2
P ( A ∩B )= 1
3
Substituting in the original equation
P ( A ∪ B ¿=3 /4+ 1/2−1/3
P ( A ∪ B ¿=11 /12
Question 2
Using the conditional probability
P(Total Children=3|students with 2 siblings)
Total number of students=30
Number of students with two siblings=9
P(Total Children=3|students with 2 siblings)=9/30 or 3/10
Question 3
Assuming each dice has 6 sides
We can therefore answer the resulting questions

(a). Possibility Space Diagram
Dice 1
1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
Dice 2 4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
(b). P(10)=2/36=1/18
(c). P(25)=1/36
(d). P(27)=0/36=0
(e). P(t)=1/9=4/36
Which value “t” has a frequency of 4 in the possibility space diagram?
6 appears four times in the possibility space diagram and it has a probability of 1/9
Then t=6
Question 4
(a). Using Bayes’ Theorem
P ( A |B ) = ( P ( B| A )∗P ( A ) }
P ( B )
Hence
P ( B| A ) = ( P ( A |B )∗P ( B ) }
P ( A )
P ( B| A ) =
2
3∗1
3
2
3

P ( B|A )= 1
2
(b). Using the notation
P ( B| A ) =P( B ∩ A)/ P ( A )
Hence
P ( B ∩ A )=P ( B|A )∗P ( A )
P ( B ∩ A )=
1
2∗2
3
P ( B ∩ A )= 2
6
Using the rule of addition
P ( B ∪ A ) =P ( B ) + P ( A ) −P ( B ∩ A )
P ( B ∪ A ) =1
3 + 2
3 − 2
6
P ( B ∪ A ) = 4
6
(b) Summing the intersection and union of A and B
A and B are exhaustive because P ( B ∩ A ) + P ( B ∪ A ) =1
Question 5
Home
D E
Early Late Early Late
0.70.3
0.20.80.70.3

(a). The probability that am late
P(L)=P(Late with E)*P(Late with D)
P(L)=0.2*0.7
P(L)=0.14
(b). The probability that am late and I take route E
P(E|L)=0.7
Question 6
(a). Possibility Tree diagram
Brian
Charlie
Ali
Origins Mungo’s K-bar Not
0.5 0.5
0.5 0.5
0.5 0.5

Find probability
(b). P(Ali and Brian meet)=0.5*0.5=0.25
(c). P(Charlie and Brian meet)=0.5*0.5=0.25
(d). P(all meet)=0
(e). P(all go to different places)=(0.5*0.5*0.5)+(0.5*0.5*0.5)=0.25
(f). P(at least 2 meet)=(0.5*0.5)+(0.5*0.5)=0.5