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Answer :
Total possible events when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)
\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)
(i) Total number of prime numbers = 3 (2, 3 and 5)
P (getting a prime number) \(= \ \frac{3}{6} \ = \ 0.5 \)
(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)
P (getting a number between 2 and 6) \( = \ \frac{3}{6} \ = \ 0.5 \)
(iii) Total number of odd numbers = 3 (1, 3 and 5)
P (getting an odd number) \( = \ \frac{3}{6} \ = \ 0.5 \)