1

### 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 = ${1 \over 3}$
2

### JEE Main 2018 (Online) 16th April Morning Slot

Let A, B and C be three events, which are pair-wise independent and $\overrightarrow E$ denotes the completement of an event E. If $P\left( {A \cap B \cap C} \right) = 0$ and $P\left( C \right) > 0,$ then $P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right]$ is equal to :
A
$P\left( {\overline A } \right) - P\left( B \right)$
B
$P\left( A \right) + P\left( {\overline B } \right)$
C
$P\left( {\overline A } \right) - P\left( {\overline B } \right)$
D
$P\left( {\overline A } \right) + P\left( {\overline B } \right)$

## Explanation

Here, $P\left( {\overline A \cap \overline B \left| C \right.} \right) = {{P\left( {\overline A \cap \overline B \cap C} \right)} \over {P\left( C \right)}}$

= ${{P\left[ {\left( {\overline {A \cup B} } \right) \cap C} \right]} \over {P\left( C \right)}}$

= ${{P\left[ {C - \left( {A \cup B} \right)} \right]} \over {P\left( C \right)}}$

= ${{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right) + P\left( {A \cap B \cap C} \right)} \over {P\left( C \right)}}$

= ${{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right)} \over {P\left( C \right)}}$ ($\because$$\left. {P\left( {A \cap B \cap C} \right) = 0} \right)$

= ${{P\left( C \right) - P\left( A \right).P(C) - P\left( B \right).P(C)} \over {P\left( C \right)}}$

[$\because$ A, B and C are independent events]

= 1 - P(A) - P(B)

= $P\left( {\overline A } \right)$ - P(B) or $P\left( {\overline B } \right)$ - P(A)
4

### JEE Main 2019 (Online) 9th January Morning Slot

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then P(X = 1) + P (X = 2) equals :
A
$25 \over 169$
B
$49\over 169$
C
$24 \over 169$
D
$52 \over 169$

## Explanation

P (X = 1) means out of two drawn cards one card is ace.

and P(X = 2) means both the drawn cards are ace.

$\therefore$  P(X = 1) = first card is ace or 2nd card is ace.

= A $-$ $+$ $-$ A

= ${4 \over {52}} \times {{48} \over {52}} + {{48} \over {52}} \times {4 \over {52}}$

= $2 \times {4 \over {52}} \times {{48} \over {52}}$

P(X = 2) =First and second both cards arc ace.

= A A

= ${4 \over {52}} \times {4 \over {52}}$

$\therefore$  P(X = 1) + P(X = 2)

= $2 \times {4 \over {52}} \times {{48} \over {52}} + {4 \over {52}} \times {4 \over {52}}$

= ${{25} \over {169}}$