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A box contains $b$ blue balls and $r$ red balls. A ball is drawn randomly from the box and is returned to the box with another ball of the same color. The probability that the second ball is drawn from the box is blue is
A.$\dfrac{b}{{r + b}}$
B.$\dfrac{{{b^2}}}{{{{\left( {r + b} \right)}^2}}}$
C.$\dfrac{{b + 1}}{{r + b + 1}}$
D.$\dfrac{{b\left( {b + 1} \right)}}{{\left( {r + b} \right)\left( {r + b + 1} \right)}}$

seo-qna
Last updated date: 13th Jun 2024
Total views: 51.6k
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Answer
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51.6k+ 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:
$P\left( B \right) = \dfrac{{{\text{Number of blue ball}}}}{{{\text{Total number of ball}}}} = \dfrac{b}{{r + b}}$]