Question

Two dice are thrown together and the total score is noted. The events E, F and G are ‘a total of 4’, ‘a total of 9 or more’, and ‘a total divisible by 5’, respectively. Calculate P(E), P(F) and P(G) and decide which pairs of events, if any, are independent.

Answer 1

Two dice are thrown together

∴ n(S) = 36

E = A total of 4 = {(2, 2), (1, 3), (3, 1)}

∴ n(E) = 3

F = A total of 9 or more

= {(3, 6), (6, 3), (5, 4), (4, 5), (5, 5), (4, 6), (6, 4), (5, 6), (6, 5), (6, 6)}

∴ n(F) = 10

G = A total divisible by 5

= {(1, 4), (4, 1), (2, 3), (3, 2), (4, 6), (6, 4), (5, 5)}

∴ n(G) = 7

Here, we see that (E ∩ F) = Φ and (E ∩ G) = Φ

And (F ∩ G) = {(4, 6), (6, 4), (5, 5)}

∴ n(F ∩ G) = 3 and (E ∩ F ∩ G) = Φ

∴ P(E) = ${\displaystyle \frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{3}{36}=\frac{1}{12}}$

P(F) = ${\displaystyle \frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{10}{36}=\frac{5}{18}}$

P(G) = ${\displaystyle \frac{\text{n}\left(\text{G}\right)}{\text{n}\left(\text{S}\right)}=\frac{7}{36}}$

P(F ∩ G) = ${\displaystyle \frac{3}{36}=\frac{1}{12}}$

And P(F) . P(G) = ${\displaystyle \frac{5}{18}\cdot \frac{7}{36}=\frac{35}{648}}$

Since, P(F ∩ G) ≠ P(F) . P(G)

Hence, there is no pair of independent events.