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Three faces of a fair dice are yellow, t...

Three faces of a fair dice are yellow, two are red and one is blue. Find the probability that the dice shows (a) yellow, (b) red and (c ) blue face.

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The correct Answer is:
`(a) (1)/(2) (b) (1)/(3) (c ) (1)/(6)`
Sample space = `{Y_(1), Y_(2), Y_(3), R_(1), R_(2), B}`, where Y stands from yellow, R for red and B for blue. ltbrltgt Now, `P(Y_(1)) =P(Y_(2)) = P(Y_(3)) = P(R_(1)) = P(R_(2)) = P(B) = (1)/(6)`.
`therefore P(Y) = P(Y_(1)) + P(Y_(2)) + P(Y_(3)) = (1)/(6) + (1)/(6) + (1)/(6) = (1)/(2)`
`P(R) = P(R_(1)) + P(R_(2)) = (1)/(6) + (1)/(6) = (1)/(3)`
And P(B) = `(1)/(6)`
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