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Suppose you alternate between throwing a normal six-sided fair die and tossing a fair coin. You start by throwing the die. What is the probability that you will see a $5$ on the die before you see tails on the coin?

  1. $\frac{1}{12}$
  2. $\frac{1}{6}$
  3. $\frac{2}{9}$
  4. $\frac{2}{7}$
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Best answer
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Answer: $\mathbf{\dfrac{2}{7}}$

Solution:

We will stop, whenever $\mathbf{\underline {5 \;\text{or Tail}}}$ is obtained


 $\therefore \mathrm P(5 \; \mathrm {\\\bf{or}} \;\text{Tail}) = \mathrm P(5) + \mathrm P(\text {Tail}) – \mathrm P(5\cap\text{ Tail}) = \frac{1}{6} + \frac{1}{2} – \left (\frac{1}{6}.\frac{1}{2} \right ) = \frac{7}{12}$

We will not stop when throwing a fair die and tossing a coin.

Its probability is given by $\mathrm P \left ( \;\text{neither}\; 5 \;\text{$\bf {or}$}\; \text{Tail}\;\right ) = \left (1 –\frac{7}{12}\right )= \frac{5}{12}$

So, the total probability of stopping by getting $5$ before Tail $= \frac{1}{6}  + \left (\frac{5}{12}\right ) \frac{1}{6}  + {\left (\frac{5}{12}\right ) }^2\frac{1}{6}  + ...$

This forms an infinite series that is in the form of GP. Therefore sum of this series is calculated by using, $\mathrm {S_{\infty}} =  \frac{a}{(1-\mathrm r)}$

Here, $\mathrm a = \frac{1}{6}, \mathrm r= \frac{5}{12}$

$\therefore$ Sum = $\dfrac{\left(\dfrac{1}{6}\right)}{\left(1-\frac{5}{12}\right)}$ = $\dfrac{\mathbf 2}{\mathbf 7}$

$\therefore \mathbf D$ is the right option.

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Here, two events [throwing the die denoted as $\mathrm{D}$ first and then tossing the coin denoted as $\mathrm{C}$] should be occurred together. Because any odd number of events like $\text{D}$, $\text{DCD}$ or $\text{DCDCD}$ will never produce the desirable outcome. So the number of events will always be even. So the even-sequence will be $$\text{DC}, \text{DCDC}, \text{DCDCDC},\cdots \tag{i}$$

The last event $\mathrm{C}$ will always be $\mathrm{T}$ (Tail). Any $\mathrm{D}$ can be the number $5$ with a condition that the following $\mathrm{C}$ will NOT be $\mathrm{T}$ (Tail) until the last $\mathrm{C}$. Besides, if any middle $\mathrm{C}$ is $\mathrm{T}$, the previous $\mathrm{D}$ can be $\{1,2,3,4,6\}$ (i.e. $1$ to $6$ except $5$).

 

So for $\text{DCDC}$, we can have cases in the form $\mathrm{5H\_T}$ where $\_$ can be replaced by any from $\{1,2,3,4,5,6\}$ [i.e. $6$ cases] and $\mathrm{{*}T5T}$ where $*$ can be replaced by any from $\{1,2,3,4,6\}$ [i.e. $5$ cases].

 

Now

Events Form of Cases Probability
$\text{DC}$ $\mathrm{5T}$ $\frac{1}{6\times2}=\frac{1}{12}$
$\text{DCDC}$ $\mathrm{5H\_T}$,  $\mathrm{{*}T5T}$ $\frac{6+5}{6\times2\times6\times2}=\frac{6+5}{12^2}$
$\text{DCDCDC}$

$\mathrm{5H\_H\_T}$,  $\mathrm{{*}T5H\_T}$,  $\mathrm{{*}T{*}T5T}$

$\frac{6^2+5\times6+5^2}{6\times2\times6\times2\times6\times2}=\frac{5^06^2+5^16^1+5^26^0}{12^3}$
$\text{DCDCDCDC}$ $\mathrm{5H\_H\_H\_T}$,  $\mathrm{{*}T5H\_H\_T}$,  $\mathrm{{*}T{*}T5H\_T}$,  $\mathrm{{*}T{*}T{*}T5T}$ $\frac{6^3+5\times6^2+5^2\times6+5^3}{6\times2\times6\times2\times6\times2\times6\times2}=\frac{5^06^3+5^16^2+5^26^1+5^36^0}{12^4}$
$\cdots$ $\cdots$ $\cdots$

 

$\therefore$ The probability of the $n^\mathrm{th}$ term of the sequence no$\mathrm{(i)}$ is

$\begin{align} P_n&=\frac{5^06^{n-1}+5^16^{n-2}+5^26^{n-3}+\cdots+5^{n-1}6^0}{12^n}\\&=\frac{6^{n-1}}{12^n} \left\{ 1+\left( \frac{5}{6} \right)+\left( \frac{5}{6} \right)^2+\left( \frac{5}{6} \right)^3+\cdots+\left( \frac{5}{6} \right)^{n-1} \right\}\\&=\frac{1}{6}\left( \frac{1}{2}\right)^n \left\{  \frac{1-\left( \frac{5}{6} \right)^n}{1-\left( \frac{5}{6} \right)}  \right\}\\&=\left( \frac{1}{2}\right)^n \left\{1-  \left(\frac{5}{6}\right)^n \right\}\\&=\left( \frac{1}{2}\right)^n-\left( \frac{5}{12}\right)^n\end{align}$

 

$\therefore$ The required probability

$\begin{align}&=P_1+P_2+P_3+\cdots\infty\\&= \left\{\left( \frac{1}{2}\right)^1-\left( \frac{5}{12}\right)^1\right\}+\left\{\left( \frac{1}{2}\right)^2-\left( \frac{5}{12}\right)^2\right\}+\left\{\left( \frac{1}{2}\right)^3-\left( \frac{5}{12}\right)^3\right\}+\cdots\infty\\&=\left\{\left( \frac{1}{2}\right)^1+\left( \frac{1}{2}\right)^2+\left( \frac{1}{2}\right)^3+\cdots\infty  \right\}-\left\{\left( \frac{5}{12}\right)^1+\left( \frac{5}{12}\right)^2+\left( \frac{5}{12}\right)^3+\cdots\infty  \right\}\\&=\frac{\frac{1}{2}}{1-\frac{1}{2}}-\frac{\frac{5}{12}}{1-\frac{5}{12}}\\&=1- \frac{5}{7}\\&=\frac{2}{7}\end{align}$

 

So the correct answer is D.

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