Question

A black and a red die are rolled.
Find the conditional probability of obtaining a sum greater 9 than given that the black die resulted in a 5.

Find the conditional probability of obtaining the sum 8 given that the red die resulted in a number less than 4.

Answer 1

Let the first observation be form the black die and second from the red die.
When two dice (one black and another red) are rolled, the sample space S = 6 x 6 = 36 (equally likely sample events)
Let E : set of events in which sum greater than 9 and F: set of events in which black die resulted in a 5
$E=\left\{\right(6,4),(4,6),(5,5),(5,6),(6,5),(6,6\left)\right\}\Rightarrow n\left(E\right)=6andF=\left\{\right(5,1),(5,2),(5,3),(5,4),(5,5),(5,6\left)\right\}\Rightarrow n\left(F\right)=6\Rightarrow E\cap F=\left\{\right(5,5),(5,6),\}\Rightarrow (E\cap F)=2$
The conditional probability of obtaining a sum greater than 9, Given that the black die resulted in a 5, is given by $P\left(\frac{E}{F}\right)$
$P\left(E\right)=\frac{Numberoffavourableoutcome}{Totalnumberofoutcomes}=\frac{6}{36}=\frac{1}{6}$
Similarly,$P\left(F\right)=\frac{6}{36}=\frac{1}{6}andP(E\cap F)=\frac{2}{36}=\frac{1}{18}\therefore P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{18}}{\frac{1}{6}}=\frac{1}{18}\times \frac{6}{1}=\frac{1}{3}$

Let the first observation be form the black die and second from the red die.
When two dice (one black and another red) are rolled, the sample space S = 6 $\times$ 6 = 36 (equally likely sample events)
Let E: set of events having 8 as the sum of the observations, F: set of events in which red die resulted in a (in any one die) number less than 4
$\therefore E=\left\{\right(2,6),(3,5),(4,4),(5,3),(6,2\left)\right\}\Rightarrow n\left(E\right)=5$

$$\left\{\begin{array}{cccccc}(1,1),& (1,2),& (1,3),& (2,1),& (2,2),& (2,3)\\ (3,1),& (3,2),& (3,3),& (4,1),& (4,2),& (4,3)\\ (5,1),& (5,2),& (5,3),& (6,1),& (6,2),& (6,3)\end{array}\right\}$$
$\Rightarrow E\cap F=\left(\right(5,3),(6,2\left)\right)n(E\cap F)=2Now,P\left(E\right)=\frac{Numberoffoverableoutcomes}{Totalnumberofoutcomes}=\frac{5}{36}Similarly,P\left(F\right)=\frac{18}{36}=\frac{1}{2}andP(E\cap F)=\frac{2}{36}=\frac{1}{18}\therefore Requiredprobability=P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{18}}{\frac{1}{2}}=\frac{1}{18}\times \frac{2}{1}=\frac{1}{9}$