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I am rolling a die, and I must roll a 3 before I roll an even number. What is the probability this happens?

 Jun 13, 2021
 #1
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The probability is 3/10.

 Jun 13, 2021
 #2
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The experiment is the following.  We keep rolling a die.  On each roll we looked at the number that appears and record it.  If the number is even, we stop the experiment; otherwise keep on rolling.

A successful event $A$ is when we performed a total of $k$ rolls for some integer $k\ge2$, and an even number (2, 4, or 6) appeared on the $k$th roll, and all prior $k-1$ rolls turned up an odd number, with 3 occurring at least once.

The probability of event $A$ seems to be hard to compute so we'll compute the probability of the complement event $\overline{A}$ and subtract it from 1 instead.

Events $\overline{A}$ has two kinds; let's call them events $B$ and $C$.  Event $B$ is when we performed a total of $k$ rolls for some positive integer $k$, and an even number (2, 4, or 6) appeared on the $k$th roll, and all prior $k-1$ rolls (if any) turned up 1 or 5.  Event $C$ is when we rolled an infinite number of times, with each roll getting a 1, 3, or 5.

The probability of event $B$ is
\begin{eqnarray*}
P(B) &=& \sum_{k=0}^\infty \bigg({\frac{1}{3}}\bigg)^k \frac{1}{2} \\
&=& \frac{1}{2}\bigg(\frac{1}{1-\frac{1}{3}}\bigg) \\
&=& \frac{1}{2}\cdot\frac{3}{2} \\
&=& \frac{3}{4}.
\end{eqnarray*}

The probability of event $C$ is $P(C) = \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\dots  = 0$.

So $P(A) = 1-P(\overline{A}) = 1-P(B)-P(C) = 1-\frac{3}{4}-0 = \frac{1}{4}$.

 Jun 13, 2021

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