Probability Rules Worksheet, Exercises of Probability and Statistics

A worksheet that contains several probability problems with solutions. The problems involve finding probabilities of events, determining if events are disjoint or independent, and using conditional probability. The problems also include Venn diagrams. The worksheet is suitable for students studying probability in university or high school.

Typology: Exercises

2021/2022

Uploaded on 05/11/2023

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Probability rules worksheet NAME:_____________________
1. If P(A) = 0.65 and P(B) = 0.32 and P(A∩B) = 0.27, find the following:
a. P(A U B) = 0.65 + 0.32 0.27 = 0.70
b. P(B|A) =
0.27 0.4154
0.65
c. Are A and B disjoint events? Why or why not?
No. P(A∩B) = 0.27 0
d. Are A and B independent events? Why or why not?
No. P(B|A) = 0.4154 P(B) = 0.32
2. If P(G) = 0.42, P(M) = 0.33 and G and M are independent, what’s the probability of G and M?
P(G∩M) = (0.42)(0.33) = 0.1386
3. If P(W) = 0.6 and P(J) = 0.34 and P(J|W) = 0.2, find the following:
a. P(W and J) = P(W J) = P(W)P(J|W) = (0.6)(0.2) = 0.12
b. P(W or J) = P(W U J) = 0.6 + 0.34 0.12 = 0.82
4. If P(Y) = 0.45 and P(L) = 0.60 and P(Y U L) = 0.78, find the following:
a. P(Y ∩ L) = 0.78 0.45 0.60 = 0.27
b. P(L|Y) =
P(L Y) 0.27 0.60
P(Y) 0.45
c. Are Y and L disjoint events? Why or why not?
No. P(Y ∩ L) = 0.27 ≠ 0
d. Are Y and L independent events? Why or why not?
Yes. P(L|Y) = 0.60 = P(L) = 0.60
5. If P(D) = 0.32, P(R) = 0.13 and D and R are disjoint, what is the probability of D or R?
P(D U R) = 0.32 + 0.13 = 0.45
6. If P(T) = 0.51 and P(B) = 0.28 and P(B|T) = 0.18, find the following:
a. P(T and B) = P(T)P(B|T) = (0.51)(0.18) = 0.0918
b. P(T or B) = 0.51 + 0.28 0.0918 = 0.6982
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Probability rules worksheet NAME:_____________________

  1. If P(A) = 0.65 and P(B) = 0.32 and P(A∩B) = 0.27, find the following: a. P(A U B) = 0.65 + 0.32 – 0.27 = 0. b. P(B|A) = 0.270.65^ 0. c. Are A and B disjoint events? Why or why not? No. P(A∩B) = 0.27 ≠ 0 d. Are A and B independent events? Why or why not? No. P(B|A) = 0.4154 ≠ P(B) = 0.
  2. If P(G) = 0.42, P(M) = 0.33 and G and M are independent, what’s the probability of G and M? P(G∩M) = (0.42)(0.33) = 0.
  3. If P(W) = 0.6 and P(J) = 0.34 and P(J|W) = 0.2, find the following: a. P(W and J) = P(W ∩ J) = P(W)∙P(J|W) = (0.6)(0.2) = 0. b. P(W or J) = P(W U J) = 0.6 + 0.34 – 0.12 = 0.
  4. If P(Y) = 0.45 and P(L) = 0.60 and P(Y U L) = 0.78, find the following: a. P(Y ∩ L) = 0.78 – 0.45 – 0.60 = 0. b. P(L|Y) =P(LP(Y)^ Y)  0.270.45^ 0. c. Are Y and L disjoint events? Why or why not? No. P(Y ∩ L) = 0.27 ≠ 0 d. Are Y and L independent events? Why or why not? Yes. P(L|Y) = 0.60 = P(L) = 0.
  5. If P(D) = 0.32, P(R) = 0.13 and D and R are disjoint, what is the probability of D or R? P(D U R) = 0.32 + 0.13 = 0.
  6. If P(T) = 0.51 and P(B) = 0.28 and P(B|T) = 0.18, find the following: a. P(T and B) = P(T)∙P(B|T) = (0.51)(0.18) = 0. b. P(T or B) = 0.51 + 0.28 – 0.0918 = 0.
  1. Suppose in a lab 24% of the mice are albino, 56% are brown, and the rest are grey.a. What is the probability that a randomly selected mouse is: i. Grey P(G) = 0. ii. Not albino P(AC) = 0. iii. Grey or Albino P(G U A) = 0. b. If the type of mouse is independent of the next what is the probability that: i. 2 randomly selected mice are both brown? P(B ∩ B) = (0.56)(0.56) = 0. ii. 2 randomly selected mice are albino then brown? P(A ∩ B) = (0.24)(0.56) = 0. iii. 2 randomly selected mice are albino and grey? P(A ∩ G) = (0.24)(0.20)*2 = 0.09 6 iv. 2 randomly selected mice are not grey? P(GC^ ∩ GC) = (0.80)(0.80) = 0. v. At least 1 out of 4 randomly selected mice is albino? P(at least 1 A out of 4) = 1 – P(AC^ ∩ AC^ ∩ AC^ ∩ AC) = 1 – (0.76)^4 = 0. vi. The first albino mouse is the 5th^ one selected? P(AC^ ∩ AC^ ∩ AC^ ∩ AC^ ∩ A) = (0.76)^4 (0.24) = 0.
  2. In the parking lot of the a large mall 64% of cars are foreign made, 12% are the color blue and 7.7% areblue and foreign made cars. a. Draw a Venn Diagram

b. What is the probability that a randomly selected car was:i. A foreign car or a blue car? P(F U B) = 0. ii. Not a foreign car and a blue car? P(FC^ ∩ B) =. iii. A foreign car given it was blue? P(F|B) = 0 0770 12^.^.^ 0 6427_._ iv. Not blue given it was not a foreign car? P(BC|FC) = 0 3170 36^.^.^ 0 8806_._

.563 .077. .

F B