Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A bag contains 4 red and 3 black balls. A second bag contains 2 red and 3 black balls. One bag is selected at random. If from the selected bag, one ball is drawn at random, then the probability that the ball drawn is red, is
MathematicsProbabilityAP EAMCETAP EAMCET 2022 (05 Jul Shift 1)
Options:
  • A $\frac{39}{70}$
  • B $\frac{41}{70}$
  • C $\frac{29}{70}$
  • D $\frac{17}{35}$
Solution:
1865 Upvotes Verified Answer
The correct answer is: $\frac{17}{35}$
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 $E_1^{\prime}, E_2$ and $A$ denote the events defined as follows
$\begin{aligned} & E_1=\text { Selecting bag I } \\ & E_2=\text { Selecting bag II }\end{aligned}$
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, $P\left(\frac{A}{E_1}\right)=$ Probability of drawing a red ball
when the first bag has been selected $=4 / 7$
$P\left(\frac{A}{E_2}\right)=$ Probability of drawing a red ball when
the second bag has been selected $=2 / 5$
$P($ red ball $)=P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)$
$=\frac{1}{2} \times \frac{4}{7}+\frac{1}{2} \times \frac{2}{5}=\frac{2}{7}+\frac{1}{5}=\frac{17}{35}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.