Conditional Probability
Suppose a card is removed from a pack of cards and kept aside.
Let B be the event that the card is a spade.
After this one card is drawn at random from the remaining 51 cards.
Let A be the event that the second card is a spade. Under the assumption that B has happened n(A) = 12
P(A/B) = 12/51
It is read as probability of A when B has taken place or probability of A under B.
P(A/B) = n(A Ç B ) / n(B)
Multiplication Theorem of Probability
If A and B are any 2 events such that
 P(A) ≠ 0
then P(A Ç B ) = P(A) . P(B/A)
 P(B) ≠ 0
then P(A Ç B) = P(B).P(A/B)
 P(B/A) = P(B)( i.e. Independent events)
then P(A/B) = P(A)
Solved Examples
 A card is drawn from a pack of cards . Given that it is a heart, what is the probability that it is not a face card?
Solution
A = The card is a heart.
B = It is not a face card.
As there are 3 face cards of heart, n(A Ç B ) = 133 = 10.
n(A) = 13
So the required probability is
P(B/A) = n(A Ç B )/n(A) = 10/13
 The probability that a student A can solve a problem is 2/3 and B can solve it 5/8 and C can solve it is 1/4. If all of them try it independently, what is the probability that the problem is solved.
Solution
Given P(A) = 2/3, P(B) = 5/8, P(C) = 1/4
So P(Á) = 1/3, P(B') = 3/8, P(C') = 3/4.
Consider the event that the problem is not solved. i.e. Á Ç B' Ç C';
So the required probability is
P(problem solved) = 1  P(problem not solved)
= 1  P(Á Ç B' Ç C';)
= 1  P(Á).P(B').P(C')
= 1  (1/3.3/8.3/4)
= 29/32
 One bag contains 4 apples and 3 mangoes and another contains 3 apples and 6 mangoes. One fruit is transferred at random from the first bag to the second and then a fruit is selected from the second bag. Find the probability that the fruit selected is an apple.
Solution
Let A be the event that an apple is transferred from the first bag and let B be the event that an apple is selected from the second bag.
So B may happen with A or Á
So B = (A Ç B ) U (A' Ç B)
A Ç B and Á Ç B are mutually exclusive.
So P(B) = P(A Ç B) + P(Á Ç B).
P( A Ç B) = P(A) . P(B/A)
= 4/7.4/10= 16/70
P(Á Ç B) = P(Á) . P( B/Á)
= 3/7.3/10
So P(B) = P( A Ç B) + P(Á Ç B)
= 16/70 + 9/70 = 25/70 = 5/14
So P(B) = 5/14
