Question & Answer
QUESTION

A bag contains 4 white and 3 black balls. Another bag contains 5 white and 2 black balls. A pair of dice is rolled. If the sum of the dice is 10, the first bag is selected. Otherwise the second bag is selected. Find the probability of drawing white ball if one ball is drawn from the selected bag at random.

ANSWER Verified Verified
Hint: - In this question firstly, we have to calculate the probability of getting sum equal to 10. Then, we have to find the probability of drawing a white ball from each bag. And finally apply the theorem of total probability for finding the drawing probability of white ball.

Complete step-by-step answer:
Total Probability Theorem: Let event C1, C2 . . . Cn forms partitions of the sample space S, where all the events have a non-zero probability of occurrence. For any event, A associated with S, the total probability of event is given by
${\text{P(A) = }}\sum\limits_{k = 0}^n {{\text{P(}}{{\text{C}}_k}){\text{P(}}\dfrac{{\text{A}}}{{{{\text{C}}_k}}})} $

The total number of possibilities on rolling a pair of dice =36
In all possibilities on rolling a pair of dice, the possibility of getting sum 10 $ = 3{\text{ \{ }}\therefore {\text{ (4,6) (5,5) (6,4) \} }}$
Let the Event is ${{\text{E}}_1}$ to get sum equal to 10 on rolling a pair of dice
Then, the probability of getting sum equal to $10$ =
$
  {\text{P(}}{{\text{E}}_1}{\text{) = }}\dfrac{3}{{36}} \\
   \Rightarrow {\text{ }}\dfrac{1}{{12}} \\
 $
Let the Event is ${{\text{E}}_2}$ to get the sum other than 10 on rolling a pair of dice
Then, the probability of getting sum other than $10$ =
$
  {\text{ P(}}{{\text{E}}_2}{\text{) = }}1 - {\text{P(}}{{\text{E}}_1}) \\
   \Rightarrow {\text{ P(}}{{\text{E}}_2}{\text{) = }}1 - {\text{ }}\dfrac{1}{{12}} \\
   \Rightarrow {\text{P(}}{{\text{E}}_2}{\text{) = }}\dfrac{{11}}{{12}} \\
    \\
  {\text{ }} \\
$
Now, Let ${\text{E}}$ be the event that white ball is drawn
Since, on rolling a pair of dice if getting sum of dice is equal to 10 then a ball is drawn from first beg

Then, probability of drawing a white ball from first beg is
$
   \Rightarrow {\text{P(}}\dfrac{{\text{E}}}{{{{\text{E}}_1}}}) \\
   \Rightarrow {\text{ P(}}\dfrac{{\text{E}}}{{{{\text{E}}_1}}}){\text{ = }}\dfrac{4}{7} \\
$
Now on rolling a pair of dice if sum of dice is other than 10 then a ball is drawn from second beg
 Then, probability of drawing a white ball from second beg is
$
   \Rightarrow {\text{P(}}\dfrac{{\text{E}}}{{{{\text{E}}_2}}}) \\
   \Rightarrow {\text{ P(}}\dfrac{{\text{E}}}{{{{\text{E}}_2}}}){\text{ = }}\dfrac{5}{7} \\
$
Now, the probability of drawing a white ball =
$
   \Rightarrow {\text{P(E) = P(}}{{\text{E}}_1}{\text{)P(}}\dfrac{{\text{E}}}{{{{\text{E}}_1}}}){\text{ + P(}}{{\text{E}}_2}{\text{)P(}}\dfrac{{\text{E}}}{{{{\text{E}}_2}}}) \\
   \Rightarrow {\text{P(E) = (}}\dfrac{1}{{12}})(\dfrac{4}{7}){\text{ + (}}\dfrac{{11}}{{12}})(\dfrac{5}{7}) \\
   \Rightarrow {\text{P(E) = }}\dfrac{{59}}{{84}} \\
$
    Hence, the probability of drawing white ball if one ball is drawn from the selected bag at
     random is $\dfrac{{59}}{{84}}$

Note:- Whenever we get this type of question the key concept of solving is we have knowledge about basic probabilities properties like ${\text{P(}}\overline {\text{E}} ) = 1 - {\text{P(E)}}$ and how to apply the total probability theorem ${\text{P(A) = }}\sum\limits_{k = 0}^n {{\text{P(}}{{\text{C}}_k}){\text{P(}}\dfrac{{\text{A}}}{{{{\text{C}}_k}}})} $ .