1

### JEE Main 2017 (Online) 9th April Morning Slot

Let E and F be two independent events. The probability that both E and F happen is ${1 \over {12}}$ and the probability that neither E nor F happens is ${1 \over {2}}$, then a value of ${{P\left( E \right)} \over {P\left( F \right)}}$ is :
A
${4 \over 3}$
B
${3 \over 2}$
C
${1 \over 3}$
D
${5 \over 12}$
2

### JEE Main 2018 (Offline)

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :
A
${3 \over 4}$
B
${3 \over 10}$
C
${2 \over 5}$
D
${1 \over 5}$

## Explanation

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

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

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

$= {2 \over 5}$
3

### JEE Main 2018 (Online) 15th April Morning Slot

A box 'A' contains $2$ white, $3$ red and $2$ black balls. Another box 'B' contains $4$ white, $2$ red and $3$ black balls. If two balls are drawn at random, without eplacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box 'B' is :
A
${9 \over {16}}$
B
${7 \over {16}}$
C
${9 \over {32}}$
D
${7 \over {8}}$

## Explanation

Probability of drawing a white ball and then a red ball

from bag B is given by ${{{}^4{C_1} \times {}^2{C_1}} \over {{}^9{C_2}}}$ = ${2 \over 9}$

Probability of drawing a white ball and then a red ball

from bag A is given by ${{{}^2{C_1} \times {}^3{C_1}} \over {{}^7{C_2}}}$ = ${2 \over 7}$

Hence, the probability of drawing a white ball and then

a red ball from bag B = ${{{2 \over 9}} \over {{2 \over 7} + {2 \over 9}}}$ = ${{2 \times 7} \over {18 + 14}}$ = ${7 \over {16}}$
4

### JEE Main 2018 (Online) 15th April Evening Slot

A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If X starts the game, and the probability of winning the game by both the players is equal, then the value of 'p' is :
A
${1 \over 5}$
B
${1 \over 3}$
C
${2 \over 5}$
D
${1 \over 4}$

## Explanation

$\therefore$ P(X getting tail) = 1 - p

P(Y getting head) = P(Y getting tail) = ${1 \over 2}$

P(X wins) = p + (1 - p)${1 \over 2}$p + (1 - p)${1 \over 2}$(1 - p)${1 \over 2}$p + ...

= ${p \over {1 - \left( {{{1 - p} \over 2}} \right)}}$

= ${{2p} \over {1 + p}}$

P(Y win) = (1 - p)${1 \over 2}$ + (1 - p)${1 \over 2}$(1 - p)${1 \over 2}$ + ...

= $\left( {{{1 - p} \over 2}} \right).{p \over {1 - \left( {{{1 - p} \over 2}} \right)}} = {{1 - p} \over {1 + p}}$

According to question,

P(X wins) = P(Y wins)

$\therefore$ ${{2p} \over {1 + p}}$ = ${{1 - p} \over {1 + p}}$

$\Rightarrow$ 3p = 1

$\Rightarrow$ p = ${1 \over 3}$