
A bag contains 12 balls out of which x are white. If one ball is drawn at random, (i) The probability that it will be a white ball is.
If 6 more white balls are put in the box. The probability of drawing a white ball will be double than that in (i). Find x.
a) 6
b) 5
c) 4
d) 3
Answer
572.1k+ views
Hint: Initially, we have a total number of balls as 12, and the number of white balls is x. So, by using the formula: $P\left( E \right)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$, find the probability of getting a white ball out of 12 balls.
Now, we added 6 white balls. So, we have the total number of balls is 18, and the number of white balls is $\left( x+6 \right)$. Again, use the formula: $P\left( E \right)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$, and find the probability of getting a white ball out of 18 balls.
Since, it is given that, probability of getting a white ball out of 18 balls is twice the probability of getting a white ball out of 12 balls, equate both the equations and solve for x.
Complete step-by-step answer:
In a box, we have:
Total number of balls = 12
Total number of white balls = x
Since, we know that: $P\left( E \right)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
So, for case (i), we get:
$P{{\left( E \right)}_{1}}=\dfrac{x}{12}......(1)$
Now, we add 6 white balls to the box.
So, we have:
Total number of balls $=12+6=18$
Total number of white balls $=\left( x+6 \right)$
So, by using the formula: $P\left( E \right)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ for case (ii), we get:
\[P{{\left( E \right)}_{2}}=\dfrac{\left( x+6 \right)}{18}......(2)\]
Since it is given that: probability of getting a white ball out of 18 balls is twice the probability of getting a white ball out of 12 balls
So, we get:
$P{{(E)}_{2}}=2\times P{{(E)}_{1}}......(3)s$
Now, substitute the value of equation (1) and equation (2) in equation (3), we get:
$\begin{align}
& \Rightarrow \dfrac{\left( x+6 \right)}{18}=\dfrac{2x}{12} \\
& \Rightarrow 12\left( x+6 \right)=36x \\
& \Rightarrow 24x=72 \\
& \Rightarrow x=3 \\
\end{align}$
Hence, x = 3
Note: As we have added 6 white balls in the box, therefore, the number of white balls increases, but simultaneously, the total number of balls also increases. Students should be careful while calculating the favorable outcomes and total outcomes.
Now, we added 6 white balls. So, we have the total number of balls is 18, and the number of white balls is $\left( x+6 \right)$. Again, use the formula: $P\left( E \right)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$, and find the probability of getting a white ball out of 18 balls.
Since, it is given that, probability of getting a white ball out of 18 balls is twice the probability of getting a white ball out of 12 balls, equate both the equations and solve for x.
Complete step-by-step answer:
In a box, we have:
Total number of balls = 12
Total number of white balls = x
Since, we know that: $P\left( E \right)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
So, for case (i), we get:
$P{{\left( E \right)}_{1}}=\dfrac{x}{12}......(1)$
Now, we add 6 white balls to the box.
So, we have:
Total number of balls $=12+6=18$
Total number of white balls $=\left( x+6 \right)$
So, by using the formula: $P\left( E \right)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ for case (ii), we get:
\[P{{\left( E \right)}_{2}}=\dfrac{\left( x+6 \right)}{18}......(2)\]
Since it is given that: probability of getting a white ball out of 18 balls is twice the probability of getting a white ball out of 12 balls
So, we get:
$P{{(E)}_{2}}=2\times P{{(E)}_{1}}......(3)s$
Now, substitute the value of equation (1) and equation (2) in equation (3), we get:
$\begin{align}
& \Rightarrow \dfrac{\left( x+6 \right)}{18}=\dfrac{2x}{12} \\
& \Rightarrow 12\left( x+6 \right)=36x \\
& \Rightarrow 24x=72 \\
& \Rightarrow x=3 \\
\end{align}$
Hence, x = 3
Note: As we have added 6 white balls in the box, therefore, the number of white balls increases, but simultaneously, the total number of balls also increases. Students should be careful while calculating the favorable outcomes and total outcomes.
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