Probability multiplication rule and Conditional probability

Question

I know that P(A|B)= probability of A given that B has occurred and it can be expressed as P(A intersection B)÷ P(B) I know about this in in single events but when we draw two things like here an urn contains 10 black and 5 white balls here we need to find probability of both balls drawn are black why conditional probability = probability of both balls drawn are black (by permutation method) ÷ probability of second event. ( conditional probability is 9/14) I want to know why it works

Answer 1

Number of black balls in the urn = 10

Number of white balls in the urn = 5

(Total number of balls in the urn = 15)

Let $A$ and $B$ denote the events that the first ball drawn is black and the second ball drawn is black, respectively.

Obviously,

$P(A) = P(\mbox{First ball drawn is black}) = {10 \over 15} = {2 \over 3}$

Since there is no replacement,

$P( B | A) = P(\mbox{Black ball in second draw given that we draw a black ball in the first draw}).$

Thus, we see that

$P(B | A) = {9 \over 14}$.

Using the multiplicative law in probability, we find that

$P(A \cap B) = P(A) P(B | A) = {2 \over 3} \times {9 \over 14} = {3 \over 7}$,

which is the required probability. $\blacksquare$