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A dice is constructed such that an outcome = even numbers 2, 4 or 6 showing up is twice as likely to show as each of the odd numbers 1, 3, and 5. What are the odds in favour of 6 showing up when the dice is rolled?

1)2 to 1

2)2 to 7

3)1 to 2

4)1 to 3

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Odds in favor of an event = (no. of favorable event) / (no. of unfavorable event).

Let the probability that odd number comes up be x.

P(1) = P(3) = P(5) = x

P(2) = P(4) = P(6) = 2x.

Odds in favor of 6 showing up = (probability of showing 6 up) / (probability of not showing 6 up)

                                                 = $\frac{2x}{2x + 2x + x + x + x}$

                                                 = $\frac{2}{7}$

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