QJEEM18Mafdb6: JEE Main 2018 Mathematics - Class 12 JEEM18Mafdb6 Solution: A bag contains 4 red and 6 black balls. A ball is drawn at…

If we follow path 1, then probability of getting 1st ball black $= {6 \over {10}}$ and probability of getting 2nd ball red when there is 4 R and 8 B balls = ${4 \over {12}}$ .

So, the probability of getting 1st ball black and 2nd ball red = ${6 \over {10}} \times {4 \over {12}}$ .

If we follow path 2, then the probability of getting 1st ball red $= {4 \over {10}}$ and probability of getting 2nd ball red when in the bag there is 6 red and 6 black balls = ${6 \over {12}}$

$\therefore\,\,\,$ Probability of getting 2nd ball as red

$= {6 \over {10}} \times {4 \over {12}} + {4 \over {10}} \times {6 \over {12}}$

$= {1 \over 5} + {1 \over 5}$

$= {2 \over 5}$