1
MCQ (Single Correct Answer)

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

P(X getting head) = p

$$ \therefore $$ 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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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)

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

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

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