Probability:
Probability is the chance of occurring an event in a random experiment.
In an experiment, the probability of happening an event E is denoted by P(E) and is defined as the ratio of number of elements in the event E to the number of elements in the sample space S.
So, P(E) = n(E)/n(S), where S is the sample space.
Example:
Find the probability of getting 2 heads when 3 coins are tossed.
Solution: sample space, S={HHH,HHT,HTH, TTH, THT,HTT, THH, TTT}, n(S) = 8.
Let E be the event of getting 2 heads, then
E={HHT, HTH, THH}, n(E)=3
So, P(E) = n(E)/n(S) = 3/8.
Conditional event:
If E1, E2 are the events of a sample space S and if E2 occurs after the occurrence of E1, then the event of occurrence of E2 after the event E1 is called conditional event which is denoted by E2|E1 or E2/E1 .
Example 1:
When three coins are tossed, the event of getting 3 heads given that there is at least one head is a conditional event.
Example 2:
When two unbiased dice are thrown, getting the sum of 8 on the dice given that one of the die showing 2 is a conditional event.
Conditional probability:
If E1 and E2 are the events of a sample space S, P(E1)≠0 then the probability of E2, after the occurrence of event E1, is called the conditional probability of the event E2 given E1 and is denoted by P(E2|E1) and we define
P(E2/E1) = P(E1∩E2)/P(E1).
Note: Since the sample space is same for both the events, the above formula can be re-written as,
P(E2/E1) = n(E1∩E2)/n(E1).
Example:
Find the probability of getting 2 tails given that there is at least one tail, when 2 unbiased coins are tossed.
Solution:
Sample space, S={HH,HT,TH,TT}
Let E1 be the event of getting at least one tail, then E1 = { HT,TH,TT}, n(E1) = 3.;
Let E2 be the event of getting 2 tails, then E1∩E2 = {TT}, n(E1∩E2) =1.
So, P(E2/E1) = n(E1∩E2)/n(E1) = 1/3
Properties of conditional probability:
1) If E is the event of a sample space S,
then P(S|E) = P(E|E) = 1
2) If E and F are any two events of a sample space S, such that P(F)≠0, then
P(E’|F) = 1-P(E|F)
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