A fair die is tossed repeatedly until a six is obtained.

Question

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 \_\_\_\_\_\_\_\_\_\_.

Answer: 12

Accepted Answer

To solve this problem, we need to compute the probabilities $a$, $b$, and $c$, and then plug those values into the expression $\frac{b+c}{a}$.

Let's begin by defining each of the variables:

$a=P(X=3)$: This is the probability that the first six appears on the third toss.

$b=P(X\geqslant 3)$: This is the probability that the first six appears on the third toss or later.

$c=P(X\geqslant 6|X>3)$: 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.

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:

Calculating $a$:

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,

$a=P(X=3)=(\frac{5}{6})\cdot (\frac{5}{6})\cdot (\frac{1}{6})$

Calculating $b$:

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 $P(X\geqslant 3)=1-P(X$P(X$P(XThus,

$b=P(X\geqslant 3)=1-P(XCalculating $c$:

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 ${(\frac{5}{6})}^{3}$. The subsequent outcomes until (and including) the fifth roll also must not be a six. So,

$c=P(X\geqslant 6|X>3)={(\frac{5}{6})}^{2}$

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.

Now we can calculate $a$, $b$, and $c$:

$a={(\frac{5}{6})}^{2}\cdot \frac{1}{6}$

$b=1-[(\frac{1}{6})+(\frac{5}{6})\cdot (\frac{1}{6})]$

$c={(\frac{5}{6})}^{2}$

Now we'll substitute to find $\frac{b+c}{a}$:

$\frac{b+c}{a}=\frac{1-[(\frac{1}{6})+(\frac{5}{6})\cdot (\frac{1}{6})]+{(\frac{5}{6})}^{2}}{{(\frac{5}{6})}^{2}\cdot \frac{1}{6}}$

Simplifying the numerator:

$1-[(\frac{1}{6})+(\frac{5}{6})\cdot (\frac{1}{6})]+{(\frac{5}{6})}^{2}$

$=1-[\frac{1}{6}+\frac{5}{36}]+\frac{25}{36}$

$=1-[\frac{6}{36}+\frac{5}{36}]+\frac{25}{36}$

$=1-\frac{11}{36}+\frac{25}{36}$

$=\frac{36}{36}-\frac{11}{36}+\frac{25}{36}$

$=\frac{50}{36}$

Now, substitute this back into the expression and solve:

$\frac{b+c}{a}=\frac{\frac{50}{36}}{{(\frac{5}{6})}^{2}\cdot \frac{1}{6}}$

$\frac{b+c}{a}=\frac{50}{36}\cdot \frac{6}{{(\frac{5}{6})}^{2}}$

$\frac{b+c}{a}=\frac{50\cdot 6}{25}$

$\frac{b+c}{a}=\frac{300}{25}$

$\frac{b+c}{a}=12$

Therefore, $\frac{b+c}{a}=12$.

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