Question

An urn contains $5$ red and $2$ green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is:
A. $\dfrac{{26}}{{49}}$
B. $\dfrac{{32}}{{49}}$
C. $\dfrac{{27}}{{49}}$
D. $\dfrac{{21}}{{49}}$

Answer 1

Hint:
In this problem, it is given that an urn contains $5$ red and $2$ green balls and we have to find the probability of drawing a red ball in the second draw. Also consider the events of drawing a red ball and replacing green ball, drawing green ball and placing a red ball, drawing a red ball in second draw and calculate the values of events.

Complete step-by-step answer:
Let,
${{\rm{E}}_1}$: Event of drawing a red ball and replacing a green ball in the bag.
${{\rm{E}}_2}$: Event of drawing a green ball and placing a red ball in the bag.
${\rm{E}}$: Event of drawing a red ball in the second draw.
It is given that an urn contains $5$ red and $2$ green balls.
$\begin{array}{c}{\rm{Thus, total \space outcomes}} = {\rm{Total \space number \space of \space balls contain \space in\space an \space urn}}\\ = 5 + 2\\ = 7\end{array}$
It is known that the probability is the ratio of the number of favorable outcomes to the total outcomes.
Now, we have to find the probability of events.
Probability of event ${{\rm{E}}_1}$: ${\rm{P}}\left( {{{\rm{E}}_1}} \right) = \dfrac{5}{7}$
Probability of event ${{\rm{E}}_2}$: ${\rm{P}}\left( {{{\rm{E}}_2}} \right) = \dfrac{2}{7}$
Probability of event $\dfrac{{\rm{E}}}{{{{\rm{E}}_1}}}$: ${\rm{P}}\left( {\dfrac{{\rm{E}}}{{{{\rm{E}}_1}}}} \right) = \dfrac{4}{7}$
Probability of event $\dfrac{{\rm{E}}}{{{{\rm{E}}_2}}}$:${\rm{P}}\left( {\dfrac{{\rm{E}}}{{{{\rm{E}}_2}}}} \right) = \dfrac{6}{7}$
Thus, the probability of event of drawing a red ball in second draw is
${\rm{P}}\left( {\rm{E}} \right) = {\rm{P}}\left( {{{\rm{E}}_1}} \right) \times {\rm{P}}\left( {\dfrac{{\rm{E}}}{{{{\rm{E}}_1}}}} \right) + {\rm{P}}\left( {{{\rm{E}}_2}} \right) \times {\rm{P}}\left( {\dfrac{{\rm{E}}}{{{{\rm{E}}_2}}}} \right)$
Substitute $\dfrac{5}{7}$ for ${\rm{P}}\left( {{{\rm{E}}_1}} \right)$, $\dfrac{4}{7}$ for ${\rm{P}}\left( {\dfrac{{\rm{E}}}{{{{\rm{E}}_1}}}} \right)$, $\dfrac{2}{7}$ for ${\rm{P}}\left( {{{\rm{E}}_2}} \right)$ and $\dfrac{6}{7}$ for ${\rm{P}}\left( {\dfrac{{\rm{E}}}{{{{\rm{E}}_2}}}} \right)$ in the above expression.
$\begin{array}{l}{\rm{P}}\left( {\rm{E}} \right) = \dfrac{5}{7} \times \dfrac{4}{7} + \dfrac{2}{7} \times \dfrac{6}{7}\\{\rm{P}}\left( {\rm{E}} \right) = \dfrac{{32}}{{49}}\end{array}$
Therefore, the probability of event of drawing a red ball in seco

Note:
Probability explains the chances of happening in an event. Here we have to calculate the probabilities of event by drawing a red ball and replacing a green ball in the bag, event of drawing a green ball and placing a red ball in the bag and then calculate the event of drawing a red ball in the second draw. Then we substitute the value of events and finally we can obtain the answer.