1

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

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

= ${{256} \over {729}}$
2

### 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

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

### 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) = ${{n\left( E \right)} \over {n\left( S \right)}}$ = ${6 \over {55}}$
4

### JEE Main 2017 (Offline)

For three events A, B and C,

P(Exactly one of A or B occurs)
= P(Exactly one of B or C occurs)
= P (Exactly one of C or A occurs) = ${1 \over 4}$
and P(All the three events occur simultaneously) = ${1 \over {16}}$.

Then the probability that at least one of the events occurs, is:
A
${7 \over {16}}$
B
${7 \over {64}}$
C
${3 \over {16}}$
D
${7 \over {32}}$

## Explanation

Given, P (A $\cap$ B $\cap$ C) = ${1 \over {16}}$

P (exactly one of A or B occurs)

= P(A) + P (B) – 2P (A $\cap$ B) = ${1 \over 4}$ .....(1)

P (Exactly one of B or C occurs)

= P(B) + P (C) – 2P (B $\cap$ C) = ${1 \over 4}$ .....(2)

P (Exactly one of C or A occurs)

= P(C) + P(A) – 2P (C $\cap$ A) = ${1 \over 4}$ .....(3)

Adding (1), (2) and (3),we get

2[ P(A) + P(B) + P (C) - P (A $\cap$ B)

- P (B $\cap$ C) - P (C $\cap$ A)] = ${3 \over 4}$

$\Rightarrow$ P(A) + P(B) + P (C) - P (A $\cap$ B)

- P (B $\cap$ C) - P (C $\cap$ A) = ${3 \over 8}$

$\therefore$ P(atleast one event occurs)

= P (A $\cup$ B $\cup$ C)

= P(A) + P(B) + P (C) - P (A $\cap$ B)

- P (B $\cap$ C) - P (C $\cap$ A) + P (A $\cap$ B $\cap$ C)

= ${3 \over 8} + {1 \over {16}}$ = ${7 \over {16}}$