Fair die: Probability of rolling $2$ before rolling $3$ or $5$

Question

Independent trials consisting of rolling a fair die are performed, what is the probability that $2$ appears before $3$ or $5?$

There are $36$ cases if we take two trials like $11 12 13 14 15 16 ..21 22 23 24 25 26..31 32 33 34 35 36$ like this . But two has occurred before , so total $6$ cases , favourable just two$(23 25)$ so ${2\over 6} ={1\over 3}$ what is wrong in this approach , answer given is $3\over 8$ .

Are you sure about $3/8$? Can you check if you posted the question accurately. This gives $p=1/3$ and not $p=3/8$ as you say. — Jimmy R., Nov 26 '15 at 16:42

Jimmy R. · Answer 1 · 2015-11-26T16:42:59.540

3

Hint: The process is a renewal process since it starts over after each roll (or terminates). So, let $X_1$ denote the result of the first draw and $p$ the required probability. Then you have that $$p=P(X_1=2)+P(X_1=1,4 \text{ or }6)p+P(X_1=3 \text{ or }5)\cdot0$$ (can you see why?).

edited Nov 26 '15 at 16:42

answered Nov 26 '15 at 16:37

Jimmy R.

35,868

score 2 · Accepted Answer · answered Nov 26 '15 at 16:58

Your answer $1/3$ is correct. The easiest way to see it is to consider the first of $2$, $3$, and $5$ to occur. (With probability $1$, one of them will eventually occur, so there is a first occurrence.) All three of $2$, $3$, and $5$ have the same probability of occurring first (of those three), so that probability must be $1/3$.

Fair die: Probability of rolling $2$ before rolling $3$ or $5$

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