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Answer :
The total number of possible outcomes (or events) = 6
\( P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes} \)
(i) The total number of faces having A on it = 2
P (getting A) \( = \ \frac{2}{6} \ = \ \frac{1}{3} \ = \ 0.33 \)
(ii) The total number of faces having D on it = 1
P (getting D) \( = \ \frac{1}{6} \ = \ 0.166 \)