Additional Theoreum
Let A and B be two events associated with a random experiment. Then
P(A U B) = P(A) + P(B) – P(A∩B)
If A and B are mutually exclusive events, then P(A U B) = P(A) + P(B) because for mutually exclusive events, P(A∩B)= 0
If A and B are two independents events, then
P(A∩B)=P(A).P(B)
Example : Two dice are rolled. What is the probability of getting an odd number in one die and getting an even number in the other die?
Total number of outcomes possible when a die is rolled, n(S) = 6 (∵ any one face out of the 6 faces)
Let A be the event of getting the odd number in one die = {1,3,5}. => n(A)= 3
P(A) =n(A)/n(S)=3/6=1/2
Let B be the event of getting an even number in the other die = {2,4, 6}. => n(B)= 3
P(B) =n(B)/n(S)=3/6=1/2
P(A∩B)=P(A).P(B)=1/2×1/2=1/4
Conditional Probability
Let A and B be two events associated with a random experiment. Then, probability of the occurrence of A given that B has already occurred is called conditional probability and denoted by P(A/B)
Example : A bag contains 5 black and 4 blue balls. Two balls are drawn from the bag one by one without replacement. What is the probability of drawing a blue ball in the second draw if a black ball is already drawn in the first draw?
Let A be the event of drawing black ball in the first draw and B be the event of drawing a blue ball in the second draw. Then, P(B/A) = Probability of drawing a blue ball in the second draw given that a black ball is already drawn in the first draw.
Total Balls = 5 + 4 = 9
Since a black ball is drawn already,
total number of balls left after the first draw = 8
total number of blue balls after the first draw = 4
P(B/A)=4/8=1/2
If n fair coins are tossed=2n
The probability of getting exactly r-number of heads when n coins are tossed =nCr/2n
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