unfair die?

Probability Level 2

A die is unfair(biased) to throw(flip) it so that the probability of getting a number is proportional to that number, i.e., $\frac {P(1)}{1} = \frac {P(2)}{2} = ... = \frac {P(6)}{6}$ . If the probability to obtain an even number when throwing the die can be written as $\frac {m}{n}$ where m,n are co-prime positive integers.

Find m + n

The answer is 11.

1 solution

Guillermo Templado
Nov 10, 2015

Let's call $\frac {P(1)}{1} = \frac {P(2)}{2} = ... = \frac {P(6)}{6} = \lambda$ . Since, P(1) + P(2) + ... + P(6) = 1 we obtain $\lambda + 2\lambda + 3\lambda + ... + 6\lambda = 1$ $\Rightarrow$ $\lambda = \frac {1}{21}$ .

Therefore, P(2) + P(4) + P(6) = $\frac {2}{21} + \frac {4}{21} + \frac {6}{21} = \frac {12}{21} = \frac{4}{7} = \frac{m}{n}$ $\Rightarrow$ m + n = 11

Upvoted!!!

Anibrata Bhattacharya - 5 years, 7 months ago

unfair die?

The answer is 11.

1 solution

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