Question

Given that the two numbers appearing on throwing two dice are different. Find the probability of the event the sum of numbers on the dice is 4.

Answer 1

When dice is thrown, number of observations in the sample space S = 6 $\times$ 6 = 36 (equally likely sample events) i.e., n (S) = 36 Let E : set of umber in which the sum fo the numbers on the dice si 4, F: set of numbers in which numbers appearing on the two dice are different
Then
E = [(1,3), (2,2), (3,1) $\Rightarrow n\left(E\right)]=3$
F = F = $$\left\{\begin{array}{ccccc}(1,2),& (1,3),& (1,4),& (1,5),& (1,6)\\ (2,1),& (2,3),& (2,4),& (2,5),& (2,6)\\ (3,1),& (3,2),& (3,4),& (3,5),& (3,6)\\ (4,1),& (4,2),& (4,3),& (4,5),& (4,6)\\ (5,1),& (5,2),& (5,3),& (5,4),& (5,6)\\ (6,1),& (6,2),& (6,3),& (6,4),& (6,5)\end{array}\right\}$$$\Rightarrow$ n(F) = 30}
n(F) = 30
Here, F contains all points of S except $\left\{\right(1,1),(2,2),(3,3),(4,4),(5,5),(6,6\left)\right\}\therefore E\cap F=\left\{\right(1,3),(3,1\left)\right\}\therefore P\left(E\right)=\frac{Totalnumberoffavourableoutcomes}{Totalnumberofoutcomes}=\frac{3}{36}=\frac{1}{12}Similarly,P\left(F\right)=\frac{30}{36}=\frac{5}{6}andP(E\cap F)=\frac{2}{36}=\frac{1}{18}$
Hence, the required probability $=P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{18}}{\frac{5}{6}}=\frac{1}{15}$