Question

Bag A contains $9$ distinct red balls & $1$ blue ball .Another Bag B contains $10$ distinct red balls. $9$ Balls are selected at random from Bag A and then transferred to Bag B. Again $9$ are selected at random from bag B and then transferred to Bag A. Then , the probability that the blue ball is still in Bag A, is

Answer 1

Hint: First step for any probability sum is to carefully read the statements and then note down the logic. Post this the student should break down the statements into smaller statements in order to simplify the sum. In This sum the balls are transferred from Bag A to Bag B and vice versa. It is necessary to identify the number of possibilities going to take place. In this case it would be only $2$. The student should also have proper knowledge of Permutations and Combinations as it may help them to calculate the probability after decoding the logic in this numerical.

Complete step-by-step answer:
It is given that Bag A consists of $9$ distinct red balls & $1$ blue ball.
Bag B consists of 10 distinct red balls .
Now we need to find the probability that the blue ball is still inside Bag A, even after all the transfers take Place.
For this sum there are $2$ cases occurring, $1st$ one being – all the $9$ balls picked up and transferred are Red in colour. $2nd$ Case is - $8$ red balls and $1$ blue ball being transferred.
Sample Space for this particular problem would be $^{10}{C_9}{ \times ^{19}}{C_9}..........(1)$
$^{10}{C_9}$- When $9$ distinct balls are transferred from Bag A to Bag B
$^{19}{C_9}$- When $9$ distinct balls are transferred from Bag B to Bag A
Numerator is the summation of $2$ cases
$1st$ Case - $9$ Balls transferred are all red in colour from Bag A to Bag B and vice Versa.
$\therefore $Probability is $^9{C_9}{ \times ^{19}}{C_9}............(2)$
$2nd$ Case – Out of $9$ Balls transferred from Bag A to Bag B and vice Versa, $8$ are red in colour & $1$ is a blue ball.
$\therefore $Probability is \[^9{C_8}{ \times ^1}{C_1}{ \times ^{18}}{C_8}{ \times ^1}{C_1}............(3)\]
$\therefore $Probability is $\dfrac{{^9{C_9}{ \times ^{19}}{C_9}{ + ^9}{C_8}{ \times ^1}{C_1}{ \times ^{18}}{C_8}{ \times ^1}{C_1}}}{{^{10}{C_9}{ \times ^{19}}{C_9}}}.............(4)$
Using the Formula of $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get the following expression
Probability is \[\dfrac{{\dfrac{{9!}}{{0!\left( 9 \right)!}} \times \dfrac{{19!}}{{9!\left( {10} \right)!}} + \dfrac{{9!}}{{8!\left( 1 \right)!}} \times 1 \times \dfrac{{18!}}{{8!\left( {10} \right)!}} \times 1}}{{\dfrac{{10!}}{{9!\left( 1 \right)!}} \times \dfrac{{19!}}{{9!\left( {10} \right)!}}}}.............(5)\]
Further simplifying we get as follows,
\[ \Rightarrow \] \[\dfrac{{1 \times \dfrac{{19 \times 18!}}{{9 \times 8! \times \left( {10} \right)!}} + 9 \times 1 \times \dfrac{{18!}}{{8!\left( {10} \right)!}} \times 1}}{{10 \times \dfrac{{19 \times 18!}}{{9 \times 8!\left( {10} \right)!}}}}.............(6)\]
Cancelling the common terms we get the following equation
\[ \Rightarrow \] \[\dfrac{{\dfrac{{19}}{9} + 9}}{{10 \times \dfrac{{19}}{9}}}.............(7)\]
Further solving we get the probability as $\dfrac{{10}}{{19}}$

The Probability of Blue ball being in Bag A is $\dfrac{{10}}{{19}}$.

Note: This sum is extremely easy to solve once the logic is clearly understood. For all the probability sums it is necessary to properly read the statements. This is because if the student doesn't understand the logic he will not be able to solve a single step. Thus when a student approaches probability numericals, he should make sure that all the basics and formulas for Permutation and combinations are clear.