Question

An urn contains ‘m’ white and ‘n’ black balls. All the balls except one, are drawn from it. The probability that the last ball left in the urn is white is,
(a)$\dfrac{m}{m+n}$
(b)$\dfrac{n}{m+n}$
(c)$\dfrac{1}{(m+n)!}$
(d)$\dfrac{mn}{(m+n)!}$

Answer 1

Hint: To solve this question, firstly we will find out the selection of q white ball from m white balls which will be left in the urn. Then, we will consider all the cases when we will draw each and every ball except one ball from the m + n balls in the urn. Then we will find the probability of that last drawn ball from the urn to be white will be equal to selecting all balls except one white ball from the urn by multiplication of all events.

Complete step-by-step solution:
Now, in question, it is given that there are ‘m’ white and ‘n’ black balls in an urn. So, there are total m + n balls in the urn.
It is asked in the question that, we have to find the probability that the last ball left in the urn is white if all the balls except one, are drawn from it.
Now, let what we do is we select 1 white ball from m white balls which are in an urn to keep it aside from all other ( m + n -1 ) balls.
We know that selecting r items from total n items is equals to $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
So, selecting 1 white ball from m white balls ${{=}^{m}}{{C}_{1}}=\dfrac{m!}{1!\left( m-1 \right)!}$….. ( I )
Now, as we have to left one white ball in the urn and take all other m + n – 1 ball out of the urn,
So, the probability of selecting all balls from except one white ball can be done is m + n -1 as balls are being taken out one by one, so we cannot take m + n – 1 ball out at a time from total m + n.
So, first we will take 1 ball out of m + n balls so probability will be $\dfrac{((m-1)+n)}{\left( m+n \right)}$, then we will take another ball out of m + n – 1 balls then probability will be $\dfrac{((m-2)+n)}{\left( m+n-1 \right)}$, then another ball out of total m + n – 2 then probability will be $\dfrac{((m-3)+n)}{\left( m+n-2 \right)}$ and so on unless and until we left with one ball in urn which is white that is in last we will take only one 1 ball out of two balls left in urn whose probability is $\dfrac{(((m-1)+n)-(m+n-2))}{\left( (m+n)-(m+n-2) \right)}$.…..( II )
As, events in ( I ) and ( II ) are mutually exclusive,
So, the probability that last drawn ball from urn to be white will be equal to selecting all balls except one white ball from the urn, that is
${{=}^{m}}{{C}_{1}}\times \dfrac{((m-1)+n)}{\left( m+n \right)}\times .....\times \dfrac{(((m-1)-(n-2))}{\left( (m+n)-(n-2) \right)}\times \dfrac{(((m-1)+n)-(n-1))}{\left( (m+n)-(n-1) \right)}\times ......\times \dfrac{(((m-1)+n)-(m+n-2))}{\left( (m+n)-(m+n-2) \right)}\times \dfrac{1}{1}$
On solving and putting $^{m}{{C}_{1}}=\dfrac{m!}{1!\left( m-1 \right)!}$ , we get
$=\dfrac{m!}{1!\left( m-1 \right)!}\times \dfrac{((m-1)+n)}{\left( m+n \right)}\times .....\times \dfrac{(((m-1)-(n-2))}{\left( (m+n)-(n-2) \right)}\times \dfrac{(((m-1)+n)-(n-1))}{\left( (m+n)-(n-1) \right)}\times ......\times \dfrac{(((m-1)+n)-(m+n-2))}{\left( (m+n)-(m+n-2) \right)}\times \dfrac{1}{1}$
On simplifying the numerical values in the brackets, we get
$=m\times \dfrac{((m-1)+n)}{\left( m+n \right)}\times .....\times \dfrac{(m+1)}{\left( m+2 \right)}\times \dfrac{(m)}{\left( m+1 \right)}\times ......\times \dfrac{(1)}{\left( 2 \right)}\times \dfrac{1}{1}$
Again, on simplifying, by cancelling out all the terms in denominator and numerator we get
\[=m\times \dfrac{1}{\left( m+n \right)}\times .....\times \dfrac{1}{1}\times \dfrac{1}{1}\times ......\times \dfrac{1}{1}\times \dfrac{1}{1}\]
Or, \[=\dfrac{m}{\left( m+n \right)}\]
Hence, option ( a ) is correct.

Note: To solve such questions, one must know what does the question actually is demanding. Also, while solving such questions, one must consider each and every case for taking out balls from the urn, and also one must know that selecting r items from total n items equal to $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. As the calculation is complex, try not to do calculation mistakes.