A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let

$a=P(X=3), b=P(X \geqslant 3)$ and $c=P(X \geqslant 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to __________.

<p>To solve this problem, we need to compute the probabilities $a$, $b$, and $c$, and then plug those values into the expression $\frac{b+c}{a}$.</p> <p>Let's begin by defining each of the variables:</p> <ul> <li>$a = P(X=3)$: This is the probability that the first six appears on the third toss.</li> <li>$b = P(X \geqslant 3)$: This is the probability that the first six appears on the third toss or later.</li> <li>$c = P(X \geqslant 6 \mid X>3)$: This is the probability that the first six appears on the sixth toss or later, given that it has not appeared in the first three tosses.</li> </ul> <p>Since we're dealing with a fair die, each side has an equal probability of $\frac{1}{6}$ of landing face up. Let's find the probabilities step by step:</p> <p><strong>Calculating $a$:</strong></p> <p> <p>The probability of rolling anything other than a six is $\frac{5}{6}$. So for the first six to show up exactly on the third roll, the sequence of rolls must be NN6, where N is anything but a six (i.e., the results of the first two rolls). Thus,</p> <p>$a = P(X=3) = \left(\frac{5}{6}\right) \cdot \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right)$</p> </p> <p><strong>Calculating $b$:</strong></p> <p> <p>For the first six to appear on the third roll or later, we can think of two cases: when the first six appears on the third roll (which we've already calculated, $a$), and when it appears after the third roll. To combine these probabilities, we can use the fact that $P(X \geqslant 3) = 1 - P(X < 3)$, where $P(X < 3)$ is the probability that the first six appears on either the first or the second roll. So we calculate the latter first:</p> <p>$ P(X < 3) = P(X=1) + P(X=2) $</p> <p>$ P(X < 3) = \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) $</p> <p>Thus,</p> <p>$ b = P(X \geqslant 3) = 1 - P(X < 3) = 1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) \right] $</p> </p> <p><strong>Calculating $c$:</strong></p> <p> <p>This is the probability that the first six appears on or after the sixth roll, given that it hasn't appeared in the first three rolls. Since $X>3$, the first three outcomes must not be a six, which occurs with probability $\left(\frac{5}{6}\right)^3$. The subsequent outcomes until (and including) the fifth roll also must not be a six. So,</p> <p>$c = P(X \geqslant 6 \mid X>3) = \left(\frac{5}{6}\right)^2$</p> <p>Notice here, we did not include the probability of rolling a six, because we are looking for the probability that we have not yet rolled a six after the fifth roll.</p> </p> <p>Now we can calculate $a$, $b$, and $c$:</p> <p> <p>$a = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}$</p> <p>$b = 1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right]$</p> <p>$c = \left(\frac{5}{6}\right)^2$</p> <p>Now we'll substitute to find $\frac{b+c}{a}$:</p> <p>$\frac{b+c}{a} = \frac{1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right] + \left(\frac{5}{6}\right)^2}{\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}}$</p> </p> <p>Simplifying the numerator:</p> <p> <p>$1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right] + \left(\frac{5}{6}\right)^2$</p> <p>$= 1 - \left[\frac{1}{6} + \frac{5}{36}\right] + \frac{25}{36}$</p> <p>$= 1 - \left[\frac{6}{36} + \frac{5}{36}\right] + \frac{25}{36}$</p> <p>$= 1 - \frac{11}{36} + \frac{25}{36}$</p> <p>$= \frac{36}{36} - \frac{11}{36} + \frac{25}{36}$</p> <p>$= \frac{50}{36}$</p> <p>Now, substitute this back into the expression and solve:</p> <p>$\frac{b+c}{a} = \frac{\frac{50}{36}}{\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}}$</p> <p>$\frac{b+c}{a} = \frac{50}{36} \cdot \frac{6}{\left(\frac{5}{6}\right)^2}$</p> <p>$\frac{b+c}{a} = \frac{50 \cdot 6}{25}$</p> <p>$\frac{b+c}{a} = \frac{300}{25}$</p> <p>$\frac{b+c}{a} = 12$</p> <p>Therefore, $\frac{b+c}{a} = 12$.</p></p> <p></p>