Question

A box contains \[12\] balls out of which \[x\] are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If \[6\] more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find \[x\].

Answer 1

Hint: First in this question we have to find the probability of drawing a black ball randomly from the box and number of black balls in the box. One relation in question is given that if \[6\] more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. By using this we are going to solve the problem.
Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes.
\[{\text{Probability = }}\dfrac{{\left( {{\text{The number of wanted outcomes}}} \right)}}{{\left( {{\text{The total possible outcomes}}} \right)}}\]

Complete step-by-step answer:
It is given that a box contains \[12\] balls out of which \[x\] are black.
One ball is drawn at random from the box.
We need to determine the probability that it will be a black ball.
Total number of balls =\[12\]
So the possible outcome of choosing one ball at random from the box, is \[^{12}{C_1}\] .
Thus the possible outcome of choosing one ball and it will be black, is \[^x{C_1}\].
The probability that one ball is taken out is a black ball
\[ \Rightarrow \dfrac{{\left( {{\text{The number of wanted outcomes}}} \right)}}{{\left( {{\text{The number of possible outcomes}}} \right)}}\]
\[ \Rightarrow \dfrac{{^x{C_1}}}{{^{12}{C_1}}}\]
 By using combination formula to solve we get,
\[ \Rightarrow \dfrac{{\dfrac{{x!}}{{1!(x - 1)!}}}}{{\dfrac{{12!}}{{1!11!}}}}\]
Simplifying the above term,
\[ \Rightarrow \dfrac{{\dfrac{{x \times (x - 1)!}}{{(x - 1)!}}}}{{\dfrac{{12 \times 11!}}{{11!}}}}\]
Cancelling the like terms in numerator and denominator to solve,
\[ \Rightarrow \dfrac{x}{{12}}\]
It is also given that, \[6\] more black balls are put in the box.
Total number of balls \[ = 12 + 6 = 18\]
So the possible outcome of choosing one ball at random from the box, is \[^{18}{C_1}\] .
Thus the possible outcome of choosing one ball and it will be black, is \[^{x + 6}{C_1}\].
The probability that one ball is taken out is a black ball
\[ \Rightarrow \dfrac{{\left( {{\text{The number of wanted outcomes}}} \right)}}{{\left( {{\text{The number of possible outcomes}}} \right)}}\]
\[ \Rightarrow \dfrac{{^{x + 6}{C_1}}}{{^{18}{C_1}}}\]
By using combination formula to solve we get,
\[ \Rightarrow \dfrac{{\dfrac{{(x + 6)!}}{{1!(x + 6 - 1)!}}}}{{\dfrac{{18!}}{{1!17!}}}}\]
Simplifying the above term,
\[ \Rightarrow \dfrac{{\dfrac{{(x + 6) \times (x + 5)!}}{{(x + 5)!}}}}{{\dfrac{{18 \times 17!}}{{17!}}}}\]
Cancelling the like terms in numerator and denominator to solve,
\[ \Rightarrow \dfrac{{x + 6}}{{18}}\]
If \[6\] more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. By using the relation in given we get,
\[ \Rightarrow \dfrac{{x + 6}}{{18}} = 2 \times \dfrac{x}{{12}}\]
Simplifying the terms to find the value of $x$,
\[ \Rightarrow \dfrac{{x + 6}}{{18}} = \dfrac{{2x}}{{12}}\]
\[ \Rightarrow \dfrac{{x + 6}}{{18}} = \dfrac{x}{6}\]
Cross multiplying the denominator of the both terms to the numerator of the both terms,
\[ \Rightarrow 6x + 36 = 18x\]
Rearranging the variable and constant terms,
\[ \Rightarrow 18x - 6x = 36\]
Subtracting the terms of $x$,
\[ \Rightarrow 12x = 36\]
Solve for $x$,
\[ \Rightarrow x = \dfrac{{36}}{{12}} = 3\]
Thus the value of \[x\] is \[3\].

Note: A combination is a grouping or subset of items.
For a combination,
\[C\left( {n,r} \right){ = ^n}{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
Where, factorial n is denoted by \[n!\] and defined by
\[n! = n(n - 1)(n - 2)(n - 4).......2.1\]