Answer:
(i) 1/6
(ii) 1/2
(iii) 1/6
Explanation:
When two fair dice are tossed,
Total possible outcomes,
n(S) = 6 × 6 = 36, ( ∵ outcome when a dice is tossed = 6 )
(i) If E = {The numbers are equal }
= { (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
⇒ n(E) = 6,
[tex]\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex]
Thus,
[tex]P(E) =\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}[/tex]
(ii) Since, even + even = even and odd + odd = even,
If F = {The sum of the numbers is even}
F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 3), (1, 5), (2, 4), (2, 6), (3, 1), (3, 5), (4, 2), (4, 6), (5, 1), (5, 3), (6, 2), (6, 4)}
⇒ n(F) = 18,
[tex]P(E) =\frac{18}{36}=\frac{1}{2}[/tex]
(iii) E∩F = E,
⇒ n(E∩F) = 6,
[tex]P(E\cap F) =\frac{1}{6}[/tex]
