Example: From the past experience with illness of his patients, a doctor has gathered the following information in a population:
5% feel that they have cancer and do have cancer;
45% feel that they have cancer and don’t have cancer;
10% do not feel they have cancer and do have it; and the remaining
40% feel that they do not have cancer and really do not have it
Solution: Denoting the events as
A, when the patient feels he has cancer, and
B, when the patient has cancer, we have
P (AB) = 0.05
P (A) = 0.5
P (B) = 0.15
The probability that a patient has cancer, given that he feels he has it, given by
P (B│A) = (P(AB))/(P(A)) = 0.05/0.5 = 0.1
The probability he feels he has cancer, given that he does have it is given by
P (A│B) = (P(AB))/(P(B)) = 0.05/0.15 = 0.33 (1) Ans.
The probability of simultaneous occurrence of 2 or more independent events is the product of the individual probabilities.
For example, in tossing 2 coins,
Probability of heads in one coin = ½
Probability of heads in another coin = ½
Thus, probability of heads in both coins = ½ x ½ = ¼ . In this case, the equation at (1) above becomes unconditional, that is, instead of P(A│B) , it will now become, P(A) only. So P(A) = P(AB)/P(B). This implies, P(AB)= P(A). P(B).
Let us consider an experiment of drawing a card from a pack of cards. Then the probability of happening of the event A: ‘The card drawn is a king’, is given by: P (A) = 4/52 = 1/13
Now, suppose that a card is drawn and we are informed that the drawn card is red. How does this information affect the likelihood of the event A?
Obviously, if the event B: ‘The card drawn is red’, has happened, the event ‘Black card’ is not possible. Hence, the probability of the event A must be computed relative to the new sample space ‘B’, which consists of 26 sample points (red cards only). Among these 26 cards, there are two (red) kings. Hence the required probability is given by:
P (A│B) = 2/26 = 1/13
