Suppose two balls are drawn sequentially without replacement from an urn containing r red and b black balls. The probability of getting a red ball on the first draw is r/(r + b).
If, however, one is told that a red ball was obtained on the first
draw, the conditional probability of getting a red ball on the second
draw is (r − 1)/(r + b − 1), because for the second draw there are r + b − 1 balls in the urn, of which r
− 1 are red. Similarly, if one is told that the first ball drawn is
black, the conditional probability of getting red on the second draw is r/(r + b − 1).
In a number of trials the relative frequency with which B occurs among those trials in which A occurs is just the frequency of occurrence of A ∩ B divided by the frequency of occurrence of A. This suggests that the conditional probability of B given A (denoted P(B|A)) should be defined by
If A denotes a red ball on the first draw and B a red ball on the second draw in the experiment of the preceding paragraph, then P(A) = r/(r + b) and
which is consistent with the “obvious” answer derived above.
Rewriting equation (4) as P(A ∩ B) = P(A)P(B|A) and adding to this expression the same expression with A replaced by Ac (“not A”) leads via equation (1) to the equality
More generally, if A1, A2,…, An are mutually exclusive events and their union is the entire sample space, so that exactly one of the Ak must occur, essentially the same argument gives a fundamental relation, which is frequently called the law of total probability:
In a number of trials the relative frequency with which B occurs among those trials in which A occurs is just the frequency of occurrence of A ∩ B divided by the frequency of occurrence of A. This suggests that the conditional probability of B given A (denoted P(B|A)) should be defined by
If A denotes a red ball on the first draw and B a red ball on the second draw in the experiment of the preceding paragraph, then P(A) = r/(r + b) and
which is consistent with the “obvious” answer derived above.
Rewriting equation (4) as P(A ∩ B) = P(A)P(B|A) and adding to this expression the same expression with A replaced by Ac (“not A”) leads via equation (1) to the equality
More generally, if A1, A2,…, An are mutually exclusive events and their union is the entire sample space, so that exactly one of the Ak must occur, essentially the same argument gives a fundamental relation, which is frequently called the law of total probability: