Answer
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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}}})} $ .
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}}})} $ .
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