1

### JEE Main 2016 (Online) 9th April Morning Slot

If A and B are any two events such that P(A) = ${2 \over 5}$ and P (A $\cap$ B) = ${3 \over {20}}$, hen the conditional probability, P(A $\left| {} \right.$(A' $\cup$ B')), where A9 denotes the complement of A, is equal to :
A
${1 \over 4}$
B
${5 \over 17}$
C
${8 \over 17}$
D
${11 \over 20}$

## Explanation

$P\left( {{A \over {A' \cup B'}}} \right)$

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

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

$\left[ \, \right.$As   $\left. {\left( {A \cap B} \right)' = A' \cap B'} \right]$

$= {{P\left( {A \cap B'} \right)} \over {P\left( {A \cap B} \right)'}}$

As   $A \cap A' = \phi$

$\therefore$   $\phi \cup \left( {A \cap B'} \right) = A \cap B'$

$= {{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( {A \cap B} \right)}}$

$\left[ \, \right.$As   $A \cap B' = A - \left( {A \cap B} \right)$

and   $\left. {\left( {A \cap B} \right)' = 1 - \left( {A \cap B} \right)} \right]$

Given  $P\left( A \right) = {2 \over 5}$

and   $P\left( {A \cap B} \right) = {3 \over {20}}$

$= {{{2 \over 5} - {3 \over {20}}} \over {1 - {3 \over {20}}}}$

$= {{{{8 - 3} \over {20}}} \over {{{17} \over {20}}}}$

$= {5 \over {17}}$
2

### JEE Main 2016 (Online) 10th April Morning Slot

An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is :
A
${{240} \over {729}}$
B
${{192} \over {729}}$
C
${{256} \over {729}}$
D
${{496} \over {729}}$

## Explanation

Probability of fail, P(F) = p

$\therefore$   Probability of success, P(S) = 2p

We know,

p + 2Pp = 1

$\Rightarrow$   p = ${1 \over 3}$

$\therefore$   Now, probability of at least 5 success in 6 trials,

P(x $\ge$ 5)

= P(x = 5) + P (x = 6)

= 6C5 ${\left( {{2 \over 3}} \right)^5}{\left( {{1 \over 3}} \right)^1}{ + ^6}{C_6}{\left( {{2 \over 3}} \right)^6}{\left( {{1 \over 3}} \right)^o}$

= ${\left( {{2 \over 3}} \right)^5}\left( {{6 \over 3} + {2 \over 3}} \right)$

= ${{256} \over {729}}$
3

### JEE Main 2017 (Offline)

A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is:
A
6
B
4
C
${6 \over {25}}$
D
${{12} \over 5}$

## Explanation

We can apply binomial probability distribution

n = 10

p = Probability of drawing a green ball = ${{15} \over {25}}$ = ${3 \over 5}$

Also q = 1 - ${3 \over 5}$ = ${2 \over 5}$

Variance = npq

= $10 \times {3 \over 5} \times {2 \over 5}$ = ${{12} \over 5}$
4

### JEE Main 2017 (Offline)

If two different numbers are taken from the set {0, 1, 2, 3, ........, 10}; then the probability that their sum as well as absolute difference are both multiple of 4, is:
A
${{12} \over {55}}$
B
${{14} \over {45}}$
C
${{7} \over {55}}$
D
${{6} \over {55}}$

## Explanation

Let A = {0, 1, 2, 3, 4, ......., 10}

Total number of ways of selecting 2 different numbers from A is

n (S) = 11C2 = 55, where 'S' denotes sample space

Let E be the given event

$\therefore$ E = {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}

$\Rightarrow$ n (E) = 6

$\therefore$ P(E) = ${{n\left( E \right)} \over {n\left( S \right)}}$ = ${6 \over {55}}$