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1

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to :
A
$${{200} \over {{6^5}}}$$
B
$${{225} \over {{6^5}}}$$
C
$${{150} \over {{6^5}}}$$
D
$${{175} \over {{6^5}}}$$

Explanation

$$\underline {} \,\,\,\underline {} \,\,\,\underline {} \,\,\underline 4 \,\,\underline 4 $$

$${1 \over {{6^2}}}\left( {{{{5^3}} \over {{6^3}}} + {{2{C_1}{{.5}^2}} \over {{6^3}}}} \right) = {{175} \over {{6^5}}}$$
2

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
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

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

$$ \because $$ $$mean \over standard\,deviation$$ = $$12\over{\sqrt3}$$ = $$4{\sqrt3}$$
3

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
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

JEE Main 2019 (Online) 11th January Morning Slot

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

Questions Asked from Probability

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