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\[Probability{\text{ }} = \;\dfrac{{favorable\;outcomes}}{{possible\;outcomes}}\]

A conditional probability is the probability of one event if another event occurred. In the “die-toss” example, the probability of event A, three dots showing, is P (A) = \[\dfrac{1}{6}\] on a single toss. But what if we know that event B, at least three dots showing, occurred? Then there are only four possible outcomes, one of which is A. The probability of A = {3} is\[\dfrac{1}{4}\], given that B = {3, 4, 5, 6} occurred. The conditional probability of A given B is written P $\left( {\dfrac{A}{B}} \right)$.

The formula of conditional probability is:

$P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A\bigcap B } \right)}}{{P(B)}}$

Event A is independent of B if the conditional probability of A given B is the same as the unconditional probability of A.

A black and a red dice are rolled.

Total possible outcomes =${6^2} = 36$………………… (1)

Let the A be an event obtaining a sum 8

$

\Rightarrow A = \left\{ {\left( {2,6} \right),\left( {3,5} \right),\left( {5,3} \right),\left( {4,4} \right),\left( {6,2} \right)} \right\} \\

\Rightarrow n\left( A \right) = 5 \\

$……………………… (2)

B be an event that the red die resulted in a number less than 4.

\[

\Rightarrow B =

\left\{

\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right), \\

\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right), \\

\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right), \\

\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right), \\

\left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right), \\

\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right) \\

\right\} \\

\Rightarrow n\left( B \right) = 18 \\

\] ……………………………….(3)

Let’s find intersect when both event A and B happens by equation (2) and (3):-

$

\Rightarrow A\bigcap B = \left\{ {\left( {5,3} \right),(6,2)} \right\} \\

\Rightarrow n\left( {A\bigcap B } \right) = 2 \\

$

Probability of $A\bigcap B $= \[\dfrac{{n\left( {A\bigcap B } \right)}}{{total\;possible\;outcomes}} = \dfrac{2}{{36}}..................(1)\]

Probability of event B = P (B) =$\dfrac{{18}}{{36}}$………………………….(2)

Using the formula mentioned in the hint of conditional probability and putting the values of equation (1) and (2).

$

P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A\bigcap B } \right)}}{{P(B)}} \\

\Rightarrow P\left( {\dfrac{A}{B}} \right) = \dfrac{{\dfrac{2}{{36}}}}{{\dfrac{{18}}{{36}}}} \\

\Rightarrow P\left( {\dfrac{A}{B}} \right) = \dfrac{1}{9} \\

$

The conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4 is $\dfrac{1}{9}$