1

### JEE Main 2017 (Online) 8th April Morning Slot

An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is :
A
${{255} \over {256}}$
B
${{127} \over {128}}$
C
${{63} \over {64}}$
D
${{1} \over {2}}$

## Explanation

An unbiased coin is tossed 8 times which is same as 8 coins tossed 1 times.

$\therefore\,\,\,$ Possible no. of out come = 28

$\therefore\,\,\,$ Sample space = 28

Here in this condition, all head or all tail out come is not acceptable.

No. of times all head can occur

(H H H H H H H H) = 1

$\therefore\,\,\,$ Probability (all head) = ${1 \over {{2^8}}}$ = ${1 \over {256}}$

No. of times all tail can occur

(T T T T T T T T) = 1

$\therefore\,\,\,$ Probability (all tail) = ${1 \over {{2^8}}}$ = ${{1 \over {256}}}$

$\therefore\,\,\,$ Required probability

= 1 $-$ (P (All head) + P (All tail))

= 1 $-$ ( ${{1 \over {256}}}$ + ${{1 \over {256}}}$)

= 1 $-$ ${{1 \over {128}}}$

= ${{127} \over {128}}$
2

### JEE Main 2017 (Online) 9th April Morning Slot

From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is :
A
${{21} \over {220}}$
B
${{3} \over {11}}$
C
${{1} \over {11}}$
D
${{2} \over {23}}$
3

### JEE Main 2017 (Online) 9th April Morning Slot

Let E and F be two independent events. The probability that both E and F happen is ${1 \over {12}}$ and the probability that neither E nor F happens is ${1 \over {2}}$, then a value of ${{P\left( E \right)} \over {P\left( F \right)}}$ is :
A
${4 \over 3}$
B
${3 \over 2}$
C
${1 \over 3}$
D
${5 \over 12}$
4

### JEE Main 2018 (Offline)

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :
A
${3 \over 4}$
B
${3 \over 10}$
C
${2 \over 5}$
D
${1 \over 5}$

## Explanation If we follow path 1, then probability of getting 1st ball black $= {6 \over {10}}$ and probability of getting 2nd ball red when there is 4 R and 8 B balls = ${4 \over {12}}$.

So, the probability of getting 1st ball black and 2nd ball red = ${6 \over {10}} \times {4 \over {12}}$.

If we follow path 2, then the probability of getting 1st ball red $= {4 \over {10}}$ and probability of getting 2nd ball red when in the bag there is 6 red and 6 black balls = ${6 \over {12}}$

$\therefore\,\,\,$ Probability of getting 2nd ball as red

$= {6 \over {10}} \times {4 \over {12}} + {4 \over {10}} \times {6 \over {12}}$

$= {1 \over 5} + {1 \over 5}$

$= {2 \over 5}$