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Total number of possible outcomes = Total number of alphabets = 26

We know,

Probability of occurrence of an event = (Total number of favorable outcomes) / (Total number of outcomes)

(i)a vowel

Favorable outcomes are a, e, i, o, u

Total number of favorable outcomes = 5

Therefore, the probability that the letter is chosen is a vowel = 5/26

(ii)a consonant

Total number of consonant = 26 – 5 = 21

Therefore, the probability that the letter is chosen is a consonant = 21/26

Let ‘S’ be the sample space.

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

⇒ n(S) = 36

Let ‘A’ and ‘B’ be the events that the sum of the numbers on the two faces is divisible by 3 and 4 respectively.

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

⇒ n(A) = 12

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

⇒ n(B) = 9

∴ P(A) =\(\frac{n(A)}{n(S)}=\frac{12}{36}\)

P(B) =\(\frac{n(A)}{n(S)}=\frac{12}{36}\)

P(B) =\(\frac{n(B)}{n(S)}=\frac{9}{36}\)

A ∩ B = {(6, 6)} ⇒ n(A ∩ B) = 1

∴ (A∩ B) =\(\frac{n(A∩ B)}{n(S)}=\frac{1}{36}\)

We know–

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

=\(\frac{13}{36}+\frac{9}{36}-\frac{1}{36}=\frac{20}{36}=\frac{5}{9}\)

∴ Probability of getting the sum multiple of3 or 4 is\(\frac{5}{9}\)

⇒ Probability of getting the sum neither a multiple of3 nor 4 = P(A ∪ B)

=\(p(\overline{A∪B})\)

= 1 – P(A∪ B)

\(= 1 - \frac{5}{9}\)

=\(\frac{4}{9}\)