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With probability we are dealing with the probability of an event happening (or not happening). An event could be anything from 'obtaining a head when flipping a coin' to 'it raining next Thursday'.
The probability that an event, A, will happen is written as P(A).
The probability that the event A, does not happen is called the complement of A and is written as A'.
As either A must or must not happen then:
P(A') = 1 − P(A)
...as probability of a certainty is equal to 1
If A and B are two events then:
A ∩ B represents the event 'both A and B occur'.
A ∪ B represents the event 'either A or B occur'.
Two events are mutually exclusive if the event of one happening excludes the other from happening. In other words, they both cannot happen simultaneously.
For exclusive events A and B then:
P(A or B) = P(A) + P(B)
This can be written in set notation as:
P(A ∪ B) = P(A) + P(B)
This can be extended for three or more exclusive events:
P(A or B or C) = P(A) + P(B) + P(C)
When a fair die is rolled find the probability of rolling a 4 or a 1.
|P(4 or 1)||= P(4) + P(1)|
|= 1/6 + 1/6|
|= 2/6 = 1/3|
|Exclusive events will involve the words 'or', 'either' or something which implies 'or'.|
|Remember 'OR' means 'add'.|
Two events are independent if the occurrence of one happening does not affect the occurrence of the other.
For independent events A and B then:
P(A and B) = P(A) × P(B)
This can be written in set notation as:
P(A ∩ B) = P(A) × P(B)
Again, this result can be extended for three or more events:
P(A and B and C) = P(A) × P(B) × P(C)
A coin is flipped at the same time as a dice is rolled. Find the probability of obtaining a head and a 5.
|P(H and 5)||= P(H) × P(5)|
|= 1/2 × 1/6|
|Independent events will involve the words 'and', 'both' or something which implies either of these.|
|Remember 'and' means 'multiply'.|
Most problems will involve a combination of exclusive and independent events. One of the best ways to answer these questions is to draw a tree diagram to cover all the arrangements.
A bag contains 8 apples. 5 are eating apples and 3 are cooking apples. If 2 apples are drawn, without replacement, find the probability that at least 1 is an eating apple.
To answer this question we will draw a tree diagram. As there are to be two picks we will draw our tree diagram in two stages. The first pick and then the second pick.
We let E, stand for eating apple and C, for cooking apple.
- In first pick, P(E) = 5/8 and P(C) = 3/8. We write the possibilities on each branch. These branches are exclusive.
- If an eating apple is picked first, then for the second pick we can either get another eating apple or a cooking apple. Now only 7 apples are left in the bag.
- However, if a cooking apple is picked first we can then pick either an eating apple or a cooking apple.
|P(at least 1 Eating apple)||= P(E and E) + P(E and C) + P(C and E)|
|= 20/56 + 15/36 + 15/36|
Note: We could have answered this question using the complement...
|P(at least 1 Eating apple)||= 1 - P(no eating apples)|
|= 1 - P (C and C)|
|= 1 - 6/56|
|= 50/56 as before|
Note: Each pair of branches sum to 1 (they are exclusive). Events between the first and second set of branches are independent events.
If two events, A and B, are not mutually exclusive then the probability that A or B will occur is given by the addition formula:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Don't panic, this just means: The probability of A or B occurring is the probability of A add the probability of B minus the probability that they both occur. This is best seen by an example...
Pick a card at random from a pack of 52 cards. Find the probability that you pick an ace or a spade.
We want P(ace or spade) or P(ace ∪ spade).
|P(ace ∪ spade)||= P(ace) + P(spade) − P(ace ∩ spade)|
|= 4/52 + 13/52 − 1/52|
|= 16/52 or 4/13|
|P(ace ∩ spade)||= P(ace and spade)|
|= 4/52 × 13/52|
|= 1/52 i.e. the card is the ace of spades|
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