Question

A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball is twice that of a red ball, find the total number of balls?

Answer 1

Hint: Let us assume that the number of blue balls is x. We know that the probability of any outcome is equal to the ratio of favorable outcomes to the total outcomes. To find the probability of any outcome we need favorable outcomes and total outcomes. Total outcomes are the possible number of ways to select a ball at random from $6+x$ number of balls. The favorable outcomes for drawing a blue ball are the possible number of ways to select a blue ball from the “x” number of balls and the favorable outcomes for drawing a red ball are the possible number of ways to select a red ball from 6 number of red balls. Now, find the respective probabilities for selecting a blue and red ball then apply the condition between their probabilities which is given in the question and hence solve the equation to get the value of x.

Complete step-by-step answer:
It is given that a bag contains 6 red balls and some blue balls. Let us assume that the number of blue balls are “x”. So, the total number of balls in a bag is equal to $6+x$.
We know that probability is equal to the ratio of favorable outcomes and total number of outcomes.
$\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$
Total outcomes are the possible number of ways to select a ball at random from $6+x$ number of balls.
We are going to select a ball at random from a $6+x$ number of balls by using a combinatorial approach.
${}^{6+x}{{C}_{1}}$
Using the relation of ${}^{n}{{C}_{1}}=n$ in the above equation we get,
$6+x$
So, the total number of outcomes is $6+x$.
Now, we are going to find the probability of drawing a blue ball from the total number of balls.
Favorable outcomes are selecting a blue ball from x number of blue balls
Selecting a blue ball from $\left( x \right)$ number of blue balls we get,
$\begin{align}
  & {}^{x}{{C}_{1}} \\
 & =x \\
\end{align}$
The probability of selecting a blue ball from the total number of balls is equal to:
$\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$
Substituting favorable outcomes as “x” and total outcomes as $6+x$ in the above equation we get,
$\text{Probability}=\dfrac{x}{6+x}$……….. Eq. (1)
The favorable outcomes for selecting a red ball from the 6 number of red balls are:
$\begin{align}
  & {}^{6}{{C}_{1}} \\
 & =6 \\
\end{align}$
The probability of drawing a red ball from the total number of balls is equal to:
$\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$
Substituting favorable outcomes as “6” and total outcomes as $6+x$ in the above equation we get,
$\text{Probability}=\dfrac{6}{6+x}$…………. Eq. (2)
It is given that the probability of drawing a blue ball is twice that of a red ball so equating the eq. (2) with the multiplication of eq. (1) by 2.
$\dfrac{6}{6+x}=2\left( \dfrac{x}{6+x} \right)$
 As you can see that $\left( 6+x \right)$ is common in the denominators of the left hand side and right hand side of the above equation so $\left( 6+x \right)$ is cancelled out from both the sides and we are left with:
$6=2x$
Dividing 2 on both the sides we get,
$3=x$
From the above, we have got the value of x as 3 so the number of blue balls is 3.
Hence, the total number of balls is equal to addition of blue and red balls which is equal to 9.

Note: There is a possibility of making a silly mistake that is instead of write the total number of balls as the addition of blue and red balls you just write the total number of balls equal to number of blue balls because after calculating the value of x in the above solution which is equal to the number of blue balls without cross checking the question you just write the number of balls as the value of x and this mistake happens when you are solving this question in last half an hour of examination so be careful of making this mistake.