An event is considered a subset of a sample space.
If we toss a die once the sample space is the set S = {1, 2, 3, 4, 5, 6} where each outcome is equally likely to happen.
If we want the event to be the number 3 on the die when it is tossed then the subset of the sample space is {3}.
It is a singleton set. Such an even this called a simple event.
If however we consider the event as 'getting a prime number' the subset of the sample space is {2, 3, 5} which has 3 elements.
Such an event is called a compound event.
If we toss a die, we can get an 'even number' or an 'odd number' but not both. So the event A of getting an even number excludes the event B of getting an odd number and vice versa. We call such events mutually exclusive events.
When we toss a coin, getting a head or getting a tail are mutually exclusive.
If we draw a card and our events are
A: getting a spade
B: getting a king
We can get the king of spades so these events are not mutually exclusive.
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Dependent and Independent Event: |
Suppose a bag contains 4 red balls and 8 black balls. Two balls are drawn one after the other without the first being replaced. Let A be the event of "getting a black ball in the first draw" and B be of "getting a red ball" in the 2nd draw. Suppose the event A occurs that is we get a black ball in the first draw then P (B) = 4/11 (4 red balls out of 12 – 1 = 11 balls).
If however A does not occur that is, instead of drawing a black ball, a red ball is drawn first, then P (B) = 3/11 (4 - 1 red balls out of 12 –1 = 11 balls)
Obviously P(B) depends on whether A occurs or not.
In such a case the events are called Dependent Events.
On the other hand, if we toss a coin twice and event A is ‘getting a head in the first toss' event B is ‘getting a tail in the second toss’, P (B) remains the same whether we get a head in the first toss or not. Such events are called Independent Events.
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Odds of an event |
Sometimes we hear people say the 'Odds of a particular team winning are 3 to 1'.
Suppose we shuffle a deck of cards what are the odds that the first card we draw is an ace?
Since there are 4 aces we know that the probability of getting an ace is |
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| Probability of not getting an ace |
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The probability of not drawing an ace is 12 times the probability of getting an ace. So the odds of getting an ace is 1 to 12 or the odds of not getting an ace is 12 to 1.
We can now form the following definition:
Suppose there are m outcomes favorable to an event and n outcomes which are unfavorable to the event in the sample space then
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Example : 21 |
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What are the odds in favor of getting a '4' in a throw of a die? What are the odds against getting a '4'?
Solution:
Number of favorable outcomes = 1
Number of unfavorable outcomes = 6 - 1 = 5 |
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| We can say that when we express odds in terms of probability then |
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Example : 22 |
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The odds in favor of an event are 4 to 5. Find the probability that it will occur
Solution:
Odds in favor of the event occurring is 4/5. Thus |
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| Probability that it will occur = 4/9. |
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Try these problems |
If P(A) = 3/5 and P(B) = 1/3 find
a. P (A or B) if A and B are mutually exclusive events
b. P(A and B) if A and B are independent events
Answer: |
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a. |
P(A) = 3/5 P(B) = 1/3
A and B are mutually exclusive events
P(A or B) = P(A ∪ B) = P(A) + P(B)
by the addition theorem of probability
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b. |
A and B are independent events
 P(A and B) = P(A ∩ B) = P(A) * P(B)
By the addition theorem of probability |
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