probabilities,probability,set operations and venn diagrams, how can we calculate probabilities without a two-set venn diagram? https://mathlibra.com/how-can-we-calculate-probabilities-without-a-two-set-venn-diagram/
How can we Calculate Probabilities without a Two-Set Venn Diagram?
The Meaning of and and or
The words and and or are very important in many areas of mathematics. We use these words in several chapters in this book. including the chapter of Probabilities. The word and is generally interpreted to mean intersection. whereas or is generally interpreted to mean union. Suppose A={1,2,3,5,6,8} and B={1,3,4,7,9,10}. The elements that belong to set Aand set B are l and 3. These are the elements in the intersection of the sets. The elements that belong to set Aor set B are 1, 2, 3, 4, 5, 6, 7, 8, 9. and 10. These are the elements in the union of the sets.
Addition rules
If A and B be any two events, then the probability of the occurrence of either event A or event B is defined by
(i) P(A∪B)=P(A)+P(B), when A and B are mutually exclusive.
(ii) P(A∪B)=P(A)+P(B)-P(A∩B), A and B are not mutually exclusive.
Note:
The above rules can be extended to more than two events.
Compound events
Before you start
You should be able to:
● add and multiply fractions.
Why do this?
Set notation can be used to describe the probability of two events occurring at the same time.
Objectives
● You will be able to use set notation to describe compound events
Key Points
● Two events are mutually exclusive when they cannot occur at the same time. For mutually exclusive events A and B:
P(A∪B)=P(A)+P(B)
● Two events are independent if one event does not affect the other event. For two independent events A and B:
P(A∩B)=P(A)×P(B)
Review
● P(A) represents the probability that the item is in set A.
● P(Ā) represents the probability that the item is not in set A.
● P(Ā) =1-P(A)
● P(A∩B) represents the probability that the item is in both set Aand set B.
● P(A∪B) represents the probability that the item is in set Aor in set B or in both sets.
● Two events are mutually exclusive when they cannot occur at the same time.
For mutually exclusive events A and B:
P(A∪B)=P(A)+P(B)
● Two events are independent if one event does not affect the other event.
For two independent events A and B:
P(A∩B)=P(A)×P(B)
Mutually Exclusive Events (Disjoint Sets)
Two events are mutually exclusive if
A∩B=Ø
This means that A and B have no elements in common.
Draw a Venn Diagram that depicts two mutually exclusive events.
📌 Example 1. M and N are mutually exclusive events.
P(M)=4/9, P(N)=⅓
Work out P(M∪N).
✍
→[M and N are mutually exclusive events. So use P(M∪N)=P(M)+P(N)]
⚠ Pay attention to every underlined singular noun in Ex2 to Ex12 just on this page!
📌 Ex2. From aset of 15 cards whose faces are numbered 1 to 15, one card is drawn at random. What is the probability that it is a multiple of 3 or 5 or both?
✍ Solution:
The sample space, if, is a set of 15 points. A={3,6,9,12,15} and so n(A)=5 B={5,10,15} and so n(B)=3
There is one point in the intersection of A and B; i.e., there is one point that belongs to both A and B, namely the number 15.
The intersection of A and B is the set of all points belonging to both A and B andis denoted by A∩B and so n(A∩B)=1.
n(A∪B)=7, because the number 15 is a multiple of both 3 and 5, so the number 15 must be included only once, regarding it as a multiple of either 3 or 5. Thus
In this case, A and B are not mutually exclusive.
A and B are mutually exclusive, if the set of points in the intersection of A and B is a null set (an empty set). That is, A∩B=Ø={}, in which case P(A∩B)=0.
📌 Ex3. A number is chosen at random from the first 25 natural numbers. What is the probability that:
(i) it is multiple of 5 or 7, (ii) it is multiple of 3 or 7.
✍ Solution:
Here, S={1, 2, 3, …, 25}.
(i) Let A= the number which is a multiple of 5={5, 10, 15, 20, 25}. Then P(A)=5/25.
Let B= the number which is a multiple of 7={7, 14, 21}. Then P(B)=3/25.
Here, A and B are mutually exclusive because A∩B=Ø={},. Therefore,
P(the number is a multiple of 5 or 7)
(ii) Let C= the number which is a multiple of 3
={3, 6, 9, 12, 15, 18, 21, 24}. Then P(C)=8/25.
Let D= the number which is a multiple of 7 ={7, 14, 21}. Then P(D) =3/25.
Here, C and D are not mutually exclusive because C∩D={21}. Then P(C∩D)=1/25. Thus,
Throwing a die and throwing a coin are independent events, since the outcome of one event does not affect the outcome of the other event. So use P(A∩B)=P(A)×P(B).
a) P(F)=⅙, b) P(H)=½, c) P(F∩H)=⅙×½=1/12
💎 Probability to get anything on Rolling or Throwing a Die
📌 Ex5. A card is drawn from a pack of 52 cards. What is the probability that it is either a spade or an ace?
✍ Solution:
Let A= the spade card and B=the ace card.
Then
Therefore,
P(the card is either a spade or an ace)
📌 Ex6. A card is drawn from a Well—shuffled deck of 52 playing cards. What is the probability that it is a queen or a heart?
✍ Q=Queen and H=Heart
💎 Probability to get a certain card, Card Withdrawal Chance, featured by 20 examples and more
📌 Ex7. The probability that a student belongs to a club is P(C)=0.4. The probability that a student works part time is P(PT)=0.5. The probability that a student belongs to a club AND works part time is P(C and PT)=0.05. What is the probability that a student belongs to a club OR works part time?
✍ Answer:
P(C∪PT)=P(C)+P(PT)-P(C∩PT)
=0.4+0.5-0.05
=0.85
📌 Ex8.
A=owns a car
B=has a pet
P(A)=0.87
P(B)=0.57
P(A and B)=0.53
What is the probability that a student owns a car OR has a pet??
✍ Answer:
P(A OR B)=P(A)+P(B)-P(A AND B)
=0.87+0.57-0.53=0.91
📌 Ex9. A survey finds that 56% of people are married. They ask the same group of people, and 67% have at least one child. If there are 41% that are married and have at least one child, what is the probability that a person in the survey is married OR has a child?
✍ Answer:
The complement rule is a way to calculate a probability based on the probability of its complement.
❶ By the definition of complement
A∪A∁=U=U
❷ Apply the probability operator
P(A∪A∁)=P(U)=1
❸ Since A and A∁ are mutually exclusive
P(A∪A∁)=P(A)+P(A∁)
❹ Hence we get P(A)=1-P(A∁)
Suppose we rolled a fair, six-sided die 10 times. Let T be the event that we roll at least 1 three. If one were to calculate T you would need to find the probability of 1 three, 2 threes, …, and 10 threes and add them all up. However, you can use the complement rule to calculate P(T).
Solution.
Let X be the times that we rolled a 3, then
If we apply the complement rule
P(T)=1-P(T∁)=1-P(X=0)
📌 Ex10. You are given that the P(A)=0.4. Calculate P(A∁).
✍ P(A∁)=1-0.4=0.6
📌 Ex11. If P(A)=0.6,P(B)=0.5 and P(A∪B=0.4), find:
(a) P(A∩B), (b) P(Ā), (c) P(B̄), (d) P(Ā∩B̄).
✍ Solution:
(a) For any two events A and B we know that
📌 Ex12. Two friends, David and Colin, frequently play golf and tennis with each other. In the long run, it has been found that David wins 3 rounds of golf out of every 5, and one game of tennis out of every 4. If they play one round of golf and one game of tennis, find the probability that David
(a) wins both,
(b) loses both,
(c) wins the round of golf only,
(d) wins either the golf or the tennis but not both.
✍ Solution:
We assume that no game ends in a draw.
Let A be the event “David wins the golf”, and so Ᾱ is the event “David loses the golf”.
P(A)+P(Ᾱ)=1
P(A)=⅗ and P(Ᾱ)=⅖
Let B be the event “David wins the tennis”, and so B̄ is the event “David loses the tennis”.
P(B)=¼ and P(B̄)=¾
On the assumption that A and Bare independent events, then
(a) P(A∩B)=P(A)⋅P(B)=⅗⋅¼=3/20
(b) P(Ᾱ∩B̄)=P(Ᾱ)⋅P(B̄)=⅖⋅¾=3/10
(c) P(A∩B̄)=P(A)⋅P(B̄)=⅗⋅¾=9/20
(d) The statement “David wins either the golf or the tennis but not both” implies that
(i) David wins the golf and loses the tennis, or
(ii) David loses the golf and wins the tennis.
(i) and (ii) are mutually exclusive.
(i) P(A∩B̄)=P(A)⋅P(B̄)=⅗⋅¾=9/20
(ii) P(Ᾱ∩B)=P(Ᾱ)⋅P(B)=⅖⋅¼=2/20
Hence the required probability
probabilities,probability,set operations and venn diagrams, how can we calculate probabilities without a two-set venn diagram? https://mathlibra.com/how-can-we-calculate-probabilities-without-a-two-set-venn-diagram/
