Question

A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

• A. $\frac{23}{42}$
• B. $\frac{19}{42}$
• C. $\frac{7}{32}$
• D. $\frac{16}{39}$

Answer 1

The correct option is B $\frac{19}{42}$

A red ball can be drawn in two mutually exclusive ways
(i) Selecting bag I and then drawing a red ball from it.
(ii) Selecting bag II and then drawing a red ball from it.
Let $E_1,E_2$ and A denote the events defined as follows:
$E_1$ = selecting bag I,
$E_2$ = selecting bag II
Since one of the two bags is selected randomly, therefore
$P\left(E_1\right)=\frac{1}{2}$ and $P\left(E_2\right)=\frac{1}{2}$.
Now, $P\left(\frac{A}{E_2}\right)$ = Probability of drawing a red ball when the first bag has been chosen $=\frac{4}{7}$
$P\left(\frac{A}{E_2}\right)$ = Probability of drawing a red ball when the second bag has been selected $=\frac{2}{6}$
Using the law of total probability, we have
$P\left(redball\right)=P\left(A\right)=P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)$
$=\frac{1}{2}\times \frac{4}{7}+\frac{1}{2}\times \frac{2}{6}=\frac{19}{42}$