Question

A bag contains an assortment of blue and red balls. If two balls are drawn at random, the probability of drawing 2 red balls is five times the probability of drawing 2 blue balls. Furthermore, the probability of drawing one ball of each color is six times the probability of drawing two blue balls. Then find the number of red and blue balls in the bag.
A. \[6, 3\]
B. \[3, 6\]
C. \[2, 7\]
D. None of these

Answer 1

Hint: Consider the number of red and blue balls. Then find the probabilities of drawing 2 red balls, 2 blue balls, and one ball of each color. After that, solve the given conditions to reach the required answer.

Formula Used: The probability of an event \[E\] is: \[P\left( E \right) = \dfrac{{Number\, of\, favourable \,outcomes}}{{Total\, number\, of\, outcomes}}\]
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Complete step by step solution:
Given:
The probability of drawing 2 red balls is five times the probability of drawing 2 blue balls.
The probability of drawing one ball of each color is six times the probability of drawing two blue balls.
Let’s consider \[x\] to be the number of red balls and \[y\] to be the number of blue balls.
Now calculate the required probabilities.
The probability of drawing two red balls is:
\[P\left( {Red} \right) = \dfrac{{{}^x{C_2}}}{{{}^{x + y}{C_2}}}\]
Simplify the above equation.
Apply combination formula.
\[P\left( {Red} \right) = \dfrac{{\dfrac{{x!}}{{2!\left( {x - 2} \right)!}}}}{{\dfrac{{\left( {x + y} \right)!}}{{2!\left( {x + y - 2} \right)!}}}}\]
\[ \Rightarrow \]\[P\left( {Red} \right) = \dfrac{{x\left( {x - 1} \right)}}{{\left( {x + y} \right)\left( {x + y - 1} \right)}}\]

The probability of drawing two blue balls is,
\[P\left( {Blue} \right) = \dfrac{{{}^y{C_2}}}{{{}^{x + y}{C_2}}}\]
Simplify the above equation.
Apply combination formula.
\[P\left( {Blue} \right) = \dfrac{{\dfrac{{y!}}{{2!\left( {y - 2} \right)!}}}}{{\dfrac{{\left( {x + y} \right)!}}{{2!\left( {x + y - 2} \right)!}}}}\]
\[ \Rightarrow \]\[P\left( {Blue} \right) = \dfrac{{y\left( {y - 1} \right)}}{{\left( {x + y} \right)\left( {x + y - 1} \right)}}\]

The probability of drawing one red and one blue ball is,
\[P\left( {one \,red \,and \,one\, blue} \right) = \dfrac{{{}^x{C_1}{}^y{C_1}}}{{{}^{x + y}{C_2}}}\]
Simplify the above equation.
Apply combination formula.
\[P\left( {one \,red \,and \,one\, blue} \right) = \dfrac{{\left( {\dfrac{{x!}}{{1!\left( {x - 1} \right)!}}} \right)\left( {\dfrac{{y!}}{{1!\left( {y - 1} \right)!}}} \right)}}{{\dfrac{{\left( {x + y} \right)!}}{{2!\left( {x + y - 2} \right)!}}}}\]
\[ \Rightarrow \]\[P\left( {one \,red \,and \,one\, blue} \right) = \dfrac{{2xy}}{{\left( {x + y} \right)\left( {x + y - 1} \right)}}\]

It is given that the probability of drawing 2 red balls is five times the probability of drawing 2 blue balls.
Then,
\[P\left( {Red} \right) = 5 \times P\left( {Blue} \right)\]
Substitute the values of the probability.
\[\dfrac{{x\left( {x - 1} \right)}}{{\left( {x + y} \right)\left( {x + y - 1} \right)}} = \dfrac{{5y\left( {y - 1} \right)}}{{\left( {x + y} \right)\left( {x + y - 1} \right)}}\]
\[ \Rightarrow \]\[x\left( {x - 1} \right) = 5y\left( {y - 1} \right)\] \[.....\left( 1 \right)\]

Also, the probability of drawing 1 ball of each colour is 6 times the probability of drawing two blue balls.
Then,
\[P\left( {one \,red \,and \,one\, blue} \right) = 6 \times P\left( {Blue} \right)\]
Substitute the values of the probability.
\[\dfrac{{2xy}}{{\left( {x + y} \right)\left( {x + y - 1} \right)}} = \dfrac{{6y\left( {y - 1} \right)}}{{\left( {x + y} \right)\left( {x + y - 1} \right)}}\]
\[ \Rightarrow \]\[2xy = 6y\left( {y - 1} \right)\]
\[ \Rightarrow \]\[x = 3\left( {y - 1} \right)\] \[.....\left( 2 \right)\]

Now substitute equation \[\left( 2 \right)\] in equation \[\left( 1 \right)\].
\[3\left( {y - 1} \right)\left( {3\left( {y - 1} \right) - 1} \right) = 5y\left( {y - 1} \right)\]
Simplify the above equation.
\[3\left( {3\left( {y - 1} \right) - 1} \right) = 5y\]
\[ \Rightarrow \]\[3\left( {3y - 3 - 1} \right) = 5y\]
\[ \Rightarrow \]\[3\left( {3y - 4} \right) = 5y\]
\[ \Rightarrow \]\[9y - 12 = 5y\]
\[ \Rightarrow \]\[4y = 12\]
Divide both sides by 4.
\[y = 3\]
Now substitute \[y = 3\] in equation \[\left( 2 \right)\].
\[x = 3\left( {3 - 1} \right)\]
\[ \Rightarrow \]\[x = 3\left( 2 \right)\]
\[ \Rightarrow \]\[x = 6\]
Therefore, the bag has 6 red balls and 3 blue balls.

Hence the correct option is A.

Note: Probability means how likely something is to happen. The probability of an event lies from 0 to 1.