Question

Consider the following statements
Statement I An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. Then, the probability that the second ball is red is $\frac{1}{2}$.
Statement II A bag contains 4 red and 4 black balls, another bay conlains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Then, the probability that the ball is drawn from the first bag is $\frac{2}{3}$.
Choose the correct option.

• A. Statement I is true; statement II is true; statement II is a correct explanation for statement I
• B. Statement I is true; statement II is true; statement II is not a correct explanation for statement I
• C. Statement I is true; statement $\Vert$ is false
• D. Statement $I$ is false; statement $\Vert$ is true

Answer 1

Answer: (B)

I.The um contains 5 red and 5 black balls.
i.e., $n(R)=5,n(B)=5$ and $n(S)=10$
Let a red ball be drawn in the first attempt
$\therefore P($ drawing a red ball $)=\frac{n(R)}{n(S)}=\frac{5}{10}=\frac{1}{2}$
If twn red halls are addef to the ulfn, then the uff contains
7 red and 5 black balls
i.e., $n(R)=7,n(B)=5$ and $n(S)=12$
$P($ drawing a red ball $)=\frac{n(R)}{n(S)}=\frac{7}{12}$
Let a black ball be drawn in the first attempt
Then, $n(R)=5,n(B)=5$ and $n(S)=10$
$\therefore P$ (drawing a black ball in the first attempt)
$=\frac{n(R)}{n(S)}=\frac{5}{10}=\frac{1}{2}$
If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.
i.e., $n(R)=5,n(B)=7$ and $n(S)=12$
$\therefore P($ drawing a red ball $)=\frac{n(R)}{n(S)}=\frac{5}{12}$
Therefore, probability of drawing second ball as red is
$=\frac{1}{2}\times \frac{7}{12}+\frac{1}{2}\times \frac{5}{12}-\frac{1}{2}\left(\frac{7}{12}+\frac{5}{12}\right)$
$=\frac{1}{2}\times 1=\frac{1}{2}$
II. Event that the ball drawn is red can be under two cases
(i) ball was drawn from first bag.
(ii) ball was drawn from second bag. Then, use Bayes' theorem
Let $E_1$ : first bag is selected, $E_2$ : second bag is selected Then, $E_1$ and $E_2$ are mutually exclusive and exhaustive. Moreover,
$P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$
Let $E$ : ball drawn is red.
$\therefore P\left(\frac{E}{E_1}\right)=P(\text{ drawing a red ball from first bag })=\frac{4}{8}=\frac{1}{2}$
$\to P\left(\frac{E}{E_2}\right)=P(\text{ drawing a red ball from second bag) }$
$=\frac{2}{8}=\frac{1}{4}$
$\therefore$ Required probability
$P\left(\frac{E_1}{E}\right)=\frac{P\left(\frac{E}{E_1}\right)P\left(E_1\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)}$
$=\frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}}=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}=\frac{\frac{1}{4}}{\frac{2+1}{8}}$
$=\frac{\frac{1}{3}}{\frac{3}{8}}=\frac{1}{4}\times \frac{8}{3}=\frac{2}{3}$