Question

In a single throw of two dice, find the probability that neither a doublet nor a total of 9 will appear.

Answer 1

Here, S = {(1, 1), (1, 2)...,(6, 5), (6, 6)}
∴ Number of possible outcomes = 6 × 6 = 36
Let E be the event where a doublet appears and F be the event where the total is 9.
i.e. E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
and F = {(3, 6), (4, 5), (5, 4), (6, 3)}

\[\therefore P\left( E \right) = \frac{6}{36} = \frac{1}{6} \text{ and }  P\left( F \right) = \frac{4}{36} = \frac{1}{9}\]

P(neither a doublet nor a total of 9) = \[P\left( \bar{E} \cap \bar{F} \right) = P\left( \bar{\left( E \cup F \right)} \right) = 1 - P\left( E \cup F \right)\]   ............(i)

The events E and F are mutually exclusive.
By addition theorem, we have:
P (E ∪ F) = P(E) + P (F)
                 = \[\frac{1}{6} + \frac{1}{9} = \frac{3 + 2}{18} = \frac{5}{18}\]

From (i), we get:

\[P\left( \bar{E} \cap \bar{F} \right) = 1 - \frac{5}{18} = \frac{18 - 5}{18} = \frac{13}{18}\]