1

### 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)
3

### 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)

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

= ${{25} \over {169}}$
4

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

An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is :
A
${{21} \over {49}}$
B
${{27} \over {49}}$
C
${{26} \over {49}}$
D
${{32} \over {49}}$

## Explanation

5 Red and 2 green balls

P(one red ball) = ${5 \over 7}$

P(one green ball) = ${2 \over 7}$

Case I :

If drawn ball is green than a red ball is added

$\left( {\matrix{ {6{\mathop{\rm Re}\nolimits} d} \cr {1\,Green} \cr } } \right)$ P (red ball) = ${6 \over 7}$

Case II :

If drawn ball is red than a green ball is added

$\left( {\matrix{ {4{\mathop{\rm Re}\nolimits} d} \cr {3\,Green} \cr } } \right)$ P (red ball) = ${4 \over 7}$

P (2nd red ball) = ${5 \over 7}$ $\times {4 \over 7} + {2 \over 7} \times {6 \over 7}$ = ${{32} \over {49}}$