CMI2019-A-6

Question

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?

• A. $\frac{1}{12}$
• B. $\frac{1}{6}$
• C. $\frac{2}{9}$
• D. $\frac{2}{7}$

Comments

Is $\frac{2}{7}$ the answer??

Answer 1

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.

Comments

@techbd123 Are you implying that this solution is not general and yours given below is? If so you are correct. This solution basically reduces the general problem to a simpler one  (as required in the given question) and that should be one's approach in exam. But if "bulb doesnt light" in the moment, one can go with the general approach and solve -- it might take some more time. I havent seen your solution carefully but looks correct in quick see. 

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Yes exactly, we doesn't go into too much of mathematical thing.

At the end of the day noone is going to use this strategy in the exam.

Also I crosschecked my solution and its perfectly fine and much easier to comprehend.

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Sir @Arjun Suresh,

Yes. I agree with you. The idea comes quickly in the exam, should be sought through. 

My solution is actually the general one which works in different variations of this problem.

Answer 2

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.

Comments

Suppose we throw a Dice and we get a 5.i.e. D happened , then why would i throw a coin ? i already got one of my desired output.

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@Satbir

Read the last of the question again. It says,

What is the probability that you will see a 5 on the die before you see tails on the coin?

It means, after getting $5$, you have to toss the coin. If you get $\mathrm{Tail}$ on the coin, you will stop. Otherwise you'll have to continue the process.

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see any similar question on keneth rosen or GO , you would get the answer.

It means, after getting a number, we have to toss the coin.

If 5 comes then we can stop.

If tail comes then we can stop.

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If 5 comes then we can stop.

Nope. You wrongly understood.
We only stop when we get $5$ and eventually $\mathrm{Tail}$ as the last event.

Suppose we got $5$ after throwing die at first.
Then we got $\mathrm{Head}$ after tossing the coin. At this moment we didn't get $\mathrm{Tail}$, so we'll continue the process again (meaning throwing the die and tossing the coin alternately) until we get $\mathrm{Tail}$ as the last one.

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Actually i think here we are using infinite trials , thats why both the answers are correct. stopping after getting 5 will not matter in long run.

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@Satbir

It's NOT about to stop after getting $5$. Actually it's about NOT taking the last event's probability ($\mathrm{Tail}$ always) into the calculation. Since as per the question, the last even should always be $\mathrm{Tail}$, hence we can take it as $100\%$ i.e. probability of $1$ instead of $\frac{1}{2}$.

In short, Jeet's solution is working because he set $\mathrm{Tail}$ as the last event always and didn't take into the calculation (meaning it's definite probability of $1$).

In this idea, after tossing the die first and then tossing the coin as the successful events (meaning $\mathrm{Tail}$ must be gotten), we can have following cases:

$1\mathrm{T}$,

$2\mathrm{T}$,

$3\mathrm{T}$,

$4\mathrm{T}$,

$5\mathrm{T}$,

$6\mathrm{T}$

Here, only favourable outcome is $5\mathrm{T}$.

So the probability in this trial $=\frac{1}{6}$. This is the idea behind Jeet's solution.

On the other hand, I took the whole cases analytically and got the probability $\frac{1}{12}$ in this trial.

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Ok , now understood.

Answer 3