Answer
64.8k+ views
Hint: First find the probability that the second ball is blue when the first ball drawn from the box is also blue and the probability that the second ball is blue when the first drawn ball is red, then find the probability that the second ball is drawn from the box is blue by adding both the probabilities.
Complete step-by-step answer:
It is given that the box contains $b$ blue balls and $r$ red balls. Then the total balls in the box are$\left( {b + r} \right)$.
There is given a condition that a ball is drawn randomly from the box and is returned to the box with another ball of the same color.
Now, assume that if the first drawn ball is blue, then the probability that the first drawn ball is blue is given as:
$P\left( B \right) = \dfrac{b}{{b + r}}$
According to the given condition the total number of blue balls in the box, after returning with another blue ball, is $\left( {b + 1} \right)$ and the number of red balls remains the same.
Then the probability that the second ball is blue$\left( {{P_{{b_1}}}} \right)$:
${P_{{b_1}}} = \dfrac{b}{{b + r}} \times \dfrac{{b + 1}}{{r + b + 1}}$ …(1)
Now, assume that if the first drawn ball is red, then the probability that the first drawn ball is red is given as:
$P\left( B \right) = \dfrac{r}{{b + r}}$
According to the given condition the total number of red balls in the box, after returning with another red ball, is $r + 1$, and the number of blue balls remains same.
Then the probability that the second ball is blue$\left( {{P_{{b_2}}}} \right)$
${P_{{b_2}}} = \dfrac{r}{{b + r}} \times \dfrac{b}{{r + b + 1}}$ …(2)
Therefore, the probability that the second drawn ball is blue is the sum of both the probability that if the first drawn ball is red$\left( {{P_{{b_1}}}} \right)$ and the first drawn ball is blue$\left( {{P_{{b_2}}}} \right)$.
$ \Rightarrow P = {P_{{b_1}}} + {P_{{b_2}}}$
Substitute $\dfrac{b}{{b + r}} \times \dfrac{{b + 1}}{{r + b + 1}}$ as the value of ${P_{{b_1}}}$ and $\dfrac{r}{{b + r}} \times \dfrac{b}{{r + b + 1}}$ as the value of${P_{{b_2}}}$ :
$ \Rightarrow P = \dfrac{b}{{b + r}} \times \dfrac{{b + 1}}{{r + b + 1}} + \dfrac{r}{{b + r}} \times \dfrac{b}{{r + b + 1}}$
Simplify the above expression:
$ \Rightarrow P = \dfrac{{b\left( {b + 1} \right) + rb}}{{\left( {b + r} \right)\left( {r + b + 1} \right)}}$
$ \Rightarrow P = \dfrac{{b\left( {r + b + 1} \right)}}{{\left( {b + r} \right)\left( {r + b + 1} \right)}}$
$ \Rightarrow P = \dfrac{b}{{\left( {b + r} \right)}}$
Therefore, the probability that the second ball is drawn from the box is blue is$\dfrac{b}{{\left( {b + r} \right)}}$.
Therefore the option A is correct.
[Note: There are $\left( {b + r} \right)$ balls in the box. The probability that the first drawn ball is red is given as:
$P\left( R \right) = \dfrac{{{\text{Number of red ball}}}}{{{\text{Total number of ball}}}} = \dfrac{r}{{r + b}}$
Similarly, the probability that the first drawn ball is blue is given as:
$P\left( B \right) = \dfrac{{{\text{Number of blue ball}}}}{{{\text{Total number of ball}}}} = \dfrac{b}{{r + b}}$]
Complete step-by-step answer:
It is given that the box contains $b$ blue balls and $r$ red balls. Then the total balls in the box are$\left( {b + r} \right)$.
There is given a condition that a ball is drawn randomly from the box and is returned to the box with another ball of the same color.
Now, assume that if the first drawn ball is blue, then the probability that the first drawn ball is blue is given as:
$P\left( B \right) = \dfrac{b}{{b + r}}$
According to the given condition the total number of blue balls in the box, after returning with another blue ball, is $\left( {b + 1} \right)$ and the number of red balls remains the same.
Then the probability that the second ball is blue$\left( {{P_{{b_1}}}} \right)$:
${P_{{b_1}}} = \dfrac{b}{{b + r}} \times \dfrac{{b + 1}}{{r + b + 1}}$ …(1)
Now, assume that if the first drawn ball is red, then the probability that the first drawn ball is red is given as:
$P\left( B \right) = \dfrac{r}{{b + r}}$
According to the given condition the total number of red balls in the box, after returning with another red ball, is $r + 1$, and the number of blue balls remains same.
Then the probability that the second ball is blue$\left( {{P_{{b_2}}}} \right)$
${P_{{b_2}}} = \dfrac{r}{{b + r}} \times \dfrac{b}{{r + b + 1}}$ …(2)
Therefore, the probability that the second drawn ball is blue is the sum of both the probability that if the first drawn ball is red$\left( {{P_{{b_1}}}} \right)$ and the first drawn ball is blue$\left( {{P_{{b_2}}}} \right)$.
$ \Rightarrow P = {P_{{b_1}}} + {P_{{b_2}}}$
Substitute $\dfrac{b}{{b + r}} \times \dfrac{{b + 1}}{{r + b + 1}}$ as the value of ${P_{{b_1}}}$ and $\dfrac{r}{{b + r}} \times \dfrac{b}{{r + b + 1}}$ as the value of${P_{{b_2}}}$ :
$ \Rightarrow P = \dfrac{b}{{b + r}} \times \dfrac{{b + 1}}{{r + b + 1}} + \dfrac{r}{{b + r}} \times \dfrac{b}{{r + b + 1}}$
Simplify the above expression:
$ \Rightarrow P = \dfrac{{b\left( {b + 1} \right) + rb}}{{\left( {b + r} \right)\left( {r + b + 1} \right)}}$
$ \Rightarrow P = \dfrac{{b\left( {r + b + 1} \right)}}{{\left( {b + r} \right)\left( {r + b + 1} \right)}}$
$ \Rightarrow P = \dfrac{b}{{\left( {b + r} \right)}}$
Therefore, the probability that the second ball is drawn from the box is blue is$\dfrac{b}{{\left( {b + r} \right)}}$.
Therefore the option A is correct.
[Note: There are $\left( {b + r} \right)$ balls in the box. The probability that the first drawn ball is red is given as:
$P\left( R \right) = \dfrac{{{\text{Number of red ball}}}}{{{\text{Total number of ball}}}} = \dfrac{r}{{r + b}}$
Similarly, the probability that the first drawn ball is blue is given as:
$P\left( B \right) = \dfrac{{{\text{Number of blue ball}}}}{{{\text{Total number of ball}}}} = \dfrac{b}{{r + b}}$]
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)