Question

A bag contains a coin of value M, and a number of other coins whose aggregate value is m. A person draws one at a time till he draws the coin M: find the value of his expectation.

Answer 1

Hint: In this question we are going to find out the expectation value of the person till he draws the coin $M$. Firstly we will find out the number of coins present in the bag and then find the probability for drawing the first, second and third coin till he gets the coin having value $M$. And then we will add all the probability to get the value of his expectation using formulas of arithmetic progression and logic. And the formula is listed below.

Forumula used: Formula of probability of successive n draws = probability of ${1^{st}}$draw $ \times $ probability of ${2^{nd}}$ draw $ \times $ probability of ${3^{rd}}$ draw $ \times $ ……………………..$ \times $ Probability of ${n^{th}}$ draw.

Sum of n terms in an AP = $\dfrac{{n(n + 1)}}{2}$

Probability of occurring the event + Probability of not occurring the event =1

Complete step-by-step solution:

First we will find out the number of coins in the bag.

According to question a bag coin of value $M$ means there is 1 coin of having $M$ value,

And there are number of coins whose aggregate value is $m$

Let assume there are $k$ coins which have an aggregate value as $m$.

So, one is the coin which has value $M$ and other is $k$ coins whose aggregate value is m

So, from the above results the total number of coins in the bag is (k+1).

Since $k$ coins having aggregate value as $m$, so using unitary method calculation the value of 1 coin,

Value of one 1 coin will be = $\dfrac{m}{k}$

Now, we are going to find out the probability for successive draws till (k+1) to get the expectation value

For ${1^{st}}$ draw the probability = probability getting coin of $M$ value = number of coin having $M$ value divided by the total number of coins in bag

We get,

Probability of ${1^{st}}$ draw =\[\dfrac{1}{k+1}\] ---------- [1]

Now, we will calculate the probability of getting coin of $M$ value at ${2^{nd}}$ draw = Probability of not getting coin of value $M$ in ${1^{st}}$ draw $ \times $ probability of getting coin of value $M$ in ${2^{nd}}$ draw

Now, the number of coins are $k$, but in first draw the number of coins will be $(k+1)$,

So we get using formula stated above

Probability of not getting coin of value $M$ will be = $(1 - \dfrac{1}{{k + 1}})$

And probability of getting coin of value $M$ will be = $\dfrac{1}{k}$

So by multiplying above two expressions we get,

Probability of getting coin of value $M$ in ${2^{nd}}$draw = \[(1 - \dfrac{1}{{k + 1}})\dfrac{1}{k}\]

Taking LCM we get,

=\[\left( {\dfrac{k}{{k + 1}}} \right)\dfrac{1}{k}\]

Dividing numerator and denominator by $k$ we get,

=\[\left( {\dfrac{1}{{k + 1}}} \right)\]     --------------- [2]

Now, we will calculate the probability of getting coin of $M$ value at ${3^{rd}}$ draw = Probability of not getting it in ${1^{st}}$ draw $ \times $ probability of not getting it in ${2^{nd}}$draw $ \times $ probability of getting coin of value $M$ in ${3^{rd}}$ draw

Now, the number of coin in${3^{rd}}$ draw will $(k-1)$, but in the first and second draw the number of coins will be $(k+1)$ and $k$ respectively

Probability of not getting coin of value $M$ will be = $(1 - \dfrac{1}{{k + 1}})$

Probability of not getting coin of value $M$ will be = \[(1 - \dfrac{1}{k})\]

Probability of getting coin of value $M$ will be = \[\dfrac{1}{{(k - 1)}}\]

So, multiplying above three expressions we get,

Probability of getting coin of value in ${3^{rd}}$ draw = \[(1 - \dfrac{1}{{k + 1}})(1 - \dfrac{1}{k})(\dfrac{1}{{k - 1}})\]

Taking LCM in above term we get,

=\[(\dfrac{k}{{k + 1}})(\dfrac{{k - 1}}{k})(\dfrac{1}{{k - 1}})\]

Dividing numerator and denominator by $(k-1)$ and $k$ we get,

=\[\dfrac{1}{{(k + 1)}}\]

Now we will calculate the probability of getting coin of value $M$ in \[{(k + 1)^{th}}\] draw = = Probability of not getting it in ${1^{st}}$ draw $ \times $ probability of not getting it in ${2^{nd}}$draw $ \times $ probability of not getting coin of value $M$ in ${3^{rd}}$ draw $ \times $……………………………………………………………..$ \times $ Probability of getting coin of value $M$ in \[{(k + 1)^{th}}\]draw

Probability of not getting coin of value $M$ in${1^{st}}$draw will be = $(1 - \dfrac{1}{{k + 1}})$

Probability of not getting coin of value $M$ in ${2^{nd}}$draw will be = \[(1 - \dfrac{1}{k})\]

Probability of not getting coin of value $M$ in ${3^{rd}}$draw will be = \[\left( {1 - \dfrac{1}{{k - 1}}} \right)\]

Probability of getting coin of value $M$ in \[{(k + 1)^{th}}\]draw will be = \[\dfrac{1}{{(k + 1)}}\]-------- [3]

So now we will calculate the expectation value = Value in ${1^{st}}$draw + Value in ${2^{nd}}$ draw + Value in ${3^{rd}}$ draw + …………………………………………..+ Value in \[{(k + 1)^{th}}\]draw

Value in ${1^{st}}$ draw = \[M(\dfrac{1}{{k + 1}})\] -------- from equation [1]

Value in ${2^{nd}}$draw =\[(\dfrac{1}{{k + 1}})\left( {M + \dfrac{m}{k}} \right)\] from above stated value of $m$ term and from equation [2]

Value in ${3^{rd}}$draw = \[(\dfrac{1}{{k + 1}})\left( {M + \dfrac{{2m}}{k}} \right)\] --------- from equation [3] 

Value in \[{(k + 1)^{th}}\] draw = \[(\dfrac{1}{{k + 1}})\left( {M + \dfrac{{km}}{k}} \right)\]

Now, adding above expression to get the expectation value 

We get,

=\[(\dfrac{1}{{k + 1}})\left( M \right) + (\dfrac{1}{{k + 1}})(M + \dfrac{m}{k}) + (\dfrac{1}{{k + 1}})(M + \dfrac{{2m}}{k}) + ............................ + (\dfrac{1}{{k + 1}})(M + \dfrac{{km}}{k})\]

Taking common \[\dfrac{1}{{(k + 1)}}\] from above equation we get,

=\[(\dfrac{1}{{k + 1}})(M + (M + \dfrac{m}{k}) + (M + \dfrac{{2m}}{k}) + ............................ + (M + \dfrac{{km}}{k}))\]

Now In above equation we know that M is repeated \[{(k + 1)^{th}}\] times so above equation becomes

=\[(\dfrac{1}{{k + 1}})(M(k + 1) + \dfrac{m}{k} + \dfrac{{2m}}{k} + ............................ + \dfrac{{km}}{k})\]

Taking \[\dfrac{m}{k}\] in above expression we get,

=\[(\dfrac{1}{{k + 1}})(M(k + 1) + \dfrac{m}{k}(1 + 2 + 3 + .................... + k))\]

Since the formula for sum of $n$ terms is stated above the sum of $k$ term will be \[\dfrac{{k(k + 1)}}{2}\]

Putting the value of sum ok $k$ terms we get,

=\[(\dfrac{1}{{k + 1}})(M(k + 1) + \dfrac{m}{k}(\dfrac{{k(k + 1)}}{2})\]

Multiplying $(k+1)$ in the expression we get,

=\[(M\dfrac{{(k + 1)}}{{(k + 1)}} + \dfrac{m}{k}(\dfrac{{}}{{(k + 1)}}\dfrac{{k(k + 1)}}{2})\]

Dividing numerator and denominator by (K+1) we get,

=\[\left( {M + \dfrac{m}{2}} \right)\]

Hence the value of his expectation will be \[\left( {M + \dfrac{m}{2}} \right)\].

Note:

By seeing the question, we must figure out what method and which formula we should use to calculate to find the desired output. For this type of question you must check given terms and what should be found out.

Use a proper formula for finding out the probability for successive events drawn and the sum of the n term in an AP and to where it should apply.