1
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 10th January Evening Slot

If the probability of hitting a target by a shooter, in any shot, is ${1 \over 3}$, then the minimum number of independent shots at the target required by him so that the probability of hitting the target atleast once is greater than ${5 \over 6}$ is -
A
4
B
6
C
5
D
3

## Explanation

$1 - {}^n{C_0}{\left( {{1 \over 3}} \right)^0}{\left( {{2 \over 3}} \right)^n} > {5 \over 6}$

${1 \over 6} > {\left( {{2 \over 3}} \right)^n}\,\, \Rightarrow \,\,0.1666 > {\left( {{2 \over 3}} \right)^n}$

${n_{\min }} = 5$
2
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 11th January Morning Slot

Two integers are selected at random from the set {1, 2, ...., 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is
A
${2 \over 5}$
B
${1 \over 2}$
C
${7 \over 10}$
D
${3 \over 5}$

## Explanation

Since sum of two numbers is even so either both are odd or both are even. Hence number of elements in reduced samples space = 5C2 + 6C2

so required probability = ${{{}^5{C_2}} \over {{}^5{C_2} + {}^6{C_2}}}$
3
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 11th January Evening Slot

Let  S = {1, 2, . . . . . ., 20}. A subset B of S is said to be "nice", if the sum of the elements of B is 203. Then the probability that a randonly chosen subset of S is "nice" is :
A
${5 \over {{2^{20}}}}$
B
${7 \over {{2^{20}}}}$
C
${4 \over {{2^{20}}}}$
D
${6 \over {{2^{20}}}}$

## Explanation

7,

1, 6

2, 5

3, 4

1, 2, 4
4
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 11th January Evening Slot

A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then $\left( {{{mean\,\,of\,X} \over {s\tan dard\,\,deviation\,\,of\,X}}} \right)$ is equal to :
A
4
B
$3\sqrt 2$
C
${{4\sqrt 3 } \over 3}$
D
$4\sqrt 3$

## Explanation

p (probability of getting white ball) = ${{30} \over {40}}$

q = ${1 \over 4}$ and n = 16

mean = np = 16.${3 \over 4}$ = 12

and standard diviation

= $\sqrt {npq}$ = $\sqrt {16.{3 \over 4}.{1 \over 4}} = \sqrt 3$

$\because$ $mean \over standard\,deviation$ = $12\over{\sqrt3}$ = $4{\sqrt3}$

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