datavorous/entrance-exam-dataset huggingface dataset:
ABC is a triangle in a plane with vertices A2, 3, 5, B-1, 3, 2 and Cλ, 5, μ . If the median through A is equally inclined to the coordinate axes, then the value of λ3+μ3+5 is

Options:
A 1130
B 1348
C 1077
D 676

The correct answer is:
1348

Solution:
Let coordinate of mid-point of BC is point D, then

D≡λ-12, 4, μ+22

And, A≡2,3,5

Therefore, direction ratios of AD are

λ-12-2, 4-3,μ+22-5

λ-52, 1,μ-82

Hence, direction cosines of AD are

λ-52λ-522+1+μ-822, 1λ-522+1+μ-822,μ-82λ-522+1+μ-822

Since, median AD is making equal angles with coordinate axes, therefore

λ-52λ-522+1+μ-822=1λ-522+1+μ-822=μ-82λ-522+1+μ-822

⇒λ-52=1=μ-82

⇒ λ=7 and μ=10

∴λ3+μ3+5=1348

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2016 (10 Apr Online)
\( A B C \) is a triangular park with \( A B=A C=100 \) metres. A vertical tower is situated at the mid-point of \( \mathrm{BC} \). If the angles of elevation of the top of the tower at, \( A \) and \( B \) are \( \cot ^{-1}(3 \sqrt{2}) \) and \( \operatorname{cosec}^{-1}(2 \sqrt{2}) \) respectively, then the height of the tower (in metres) is

Options:
A \( \frac{100}{3 \sqrt{3}} \)
B \( 20 \)
C \( 25 \)
D \( 10 \sqrt{5} \)

The correct answer is:
\( 20 \)

Solution:
Let AD=y & DE=h (Tower)

∴ y2=1002-x2

Now cotθ=32

⇒tanθ=132=hy

⇒y=32h

⇒y2=1002-x2=18h2 …1

⇒tan∅=17=hx

⇒x=h7 …2

Form 1 & 2 1002-7h2=18h2

⇒ 25h2=1002⇒h=20

Topic: Mathematics, Heights and Distances, JEE Main
$\mathbf{a} \times(\mathbf{b} \times \mathbf{c})$ is equal to

Options:
A $(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{a}) \mathbf{b}$
B $(\mathbf{a} \cdot \mathbf{c}) \mathbf{a}-(\mathbf{b} \cdot \mathbf{c}) \mathbf{a}$
C $(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}$
D $(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}-(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}$

The correct answer is:
$(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}$

Solution:
It is a fundamental concept.
(a.c)b-(a.b)c.
This is a fundamental concept of vector calculus. The vector triple product is a binary operation that takes three vectors and produces a new vector. It is also known as the cross product of a vector and a cross product.
The formula for the vector triple product is as follows:
$a \times(b \times c)=(a . c) b-(a . b) c$

Topic: Mathematics, Vector Algebra, JEE Main
$\int \frac{\sin x d x}{(a+b \cos x)^2}=$

Options:
A $\frac{1}{b}(a+b \cos x)+c$
B $\frac{1}{b(a+b \cos x)}+c$
C $\frac{1}{b} \log (a+b \cos x)+c$
D None of these

The correct answer is:
$\frac{1}{b(a+b \cos x)}+c$

Solution:
$\begin{aligned} & \text { Put } a+b \cos x=t \Rightarrow d x=-\frac{d t}{b \sin x} \text {, then } \\ & \int \frac{\sin x}{(a+b \cos x)^2} d x=-\frac{1}{b} \int \frac{1}{t^2} d t=\frac{1}{b} \frac{1}{t}+c \\ & =\frac{1}{b(a+b \cos x)}+c . \\ & \end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\mathrm{A}, \mathrm{B}, \mathrm{C}$ try to hit a target simultaneously but independently. Their respective probabilities of hitting the targets are $\frac{3}{4}, \frac{1}{2}, \frac{5}{8}$. The probability that the target is hit by A or B but not by $\mathrm{C}$ is :

Options:
A
$21 / 64$

B
$7 / 8$

C
$7 / 32$

D
$9 / 64$

The correct answer is:
$21 / 64$

Solution:
$\mathrm{P}(\mathrm{A}$ or $\mathrm{B}$ but not by $\mathrm{C})=\mathrm{P}((\mathrm{A} \cup \mathrm{B}) \cap \overline{\mathrm{C}})$
$$
\begin{aligned}
& =\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \times \mathrm{P}(\overline{\mathrm{C}}) \\
& =[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})] \times \mathrm{P}(\overline{\mathrm{C}}) \\
& = \\
& {\left[\frac{3}{4}+\frac{1}{2}-\frac{3}{4} \times \frac{1}{2}\right] \times \frac{3}{8}=\left(\frac{6+4-3}{8}\right) \times \frac{3}{8}=\frac{21}{64}}
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2013 (23 Apr Online)
$\quad A B$ is a potentiometer wire (figure). If the value of $R$ is increased, in which direction will the balance point $\mathrm{J}$ shift?

Options:

None

Solution:
If $\mathrm{R}$ increases then, the current in main circuit decreases which in turn, decreases the potential difference across $\mathrm{AB}$ and hence potential gradient $(\mathrm{K})$ across $\mathrm{AB}$ decreases. Since, at neutral point, for given emf of cell, I increases as potential gradient $(\mathrm{K})$ across $\mathrm{AB}$ has decreased because.
$$
\mathrm{E}=\mathrm{kI} \quad\left[\because \mathrm{K}=\frac{\mathrm{V}}{(\mathrm{AB})}\right]
$$
So balance potential accross $\mathrm{AB}$ equal to potential of secondary circuit the length $(\mathrm{AJ})$ must be longer then earlier.
Thus, with increase of 1 , the balance point neutral point will shift towards B.

Topic: Physics, Current Electricity
$A B$ is a variable chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. If $A B$ subtends a right angle at the origin $O$, then $\frac{1}{OA^2}+\frac{1}{OB^2}$ equals to

Options:
A $\frac{1}{a^2}+\frac{1}{b^2}$
B $\frac{1}{a^2}-\frac{1}{b^2}$
C $a^2+b^2$
D $a^2-b^2$

The correct answer is:
$\frac{1}{a^2}+\frac{1}{b^2}$

Solution:

Topic: Mathematics, Ellipse, JEE Main
$A B$ is a vertical pole with $B$ at the ground level and $A$ at the top. $A$ man finds that the angle of elevation of the point $A$ from a certain point $C$ on the ground is $60^{\circ}$. He moves away from the pole along the line $B C$ to a point $D$ such that $C D=7 \mathrm{~m}$. From $D$ the angle of elevation of the point $A$ is $45^{\circ}$. Then the height of the pole is

Options:
A
$\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}-1} m$

B
$\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}+1) m$

C
$\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}-1) \mathrm{m}$

D
$\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}+1}$

The correct answer is:
$\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}+1) m$

Solution:
$$
\begin{aligned}
& \mathrm{BD}=\mathrm{AB}=7+x \\
& \text { Also } \mathrm{AB}=\mathrm{x} \tan 60^{\circ}=x \sqrt{3} \\
& \therefore x \sqrt{3}=7+x \\
& \mathrm{x}=\frac{7}{\sqrt{3}-1} \\
& \mathrm{AB}=\frac{7 \sqrt{3}}{2}(\sqrt{3}+1)
\end{aligned}
$$

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2008
AB is a vertical tower. The point A is on the ground and C is the middle point of AB. The part CB subtend an angle Î± at a point P on the ground. If AP=n AB, then the correct relation is

Options:
A n=n2+1tan â¡α
B n=2n2-1tan â¡α
C n2=2n2+1tan â¡α
D n=2n2+1tan â¡Î±

The correct answer is:
n=2n2+1tan â¡Î±

Solution:
In âPAB, tanâ¡Î²=ABAP
In âPAC,tan â¡Î¸=ACAP
â´ tan â¡Î±=tanâ¡Î²-Î¸
=tan â¡Î²-tan â¡Î¸1+tan â¡Î² tan â¡Î¸
= ABAP-ACAP1+ABAPâACAP ...(i)
âµ AP=n AB=n2AC (âµC is the mid point of BA)
From Eq. (i), we get
tan â¡Î±=1n-12n1+12n2=n2n2+1
â n=2n2+1tan â¡Î±

Topic: Mathematics, Heights and Distances, JEE Main
A→BReactantProduct

If formation of compound B follows the first order of kinetics and after 70 minutes the concentration of A was found to be half of its initial concentration. Then the rate constant of the reaction is x×10-6 s-1. The value of x is____

Options:

The correct answer is:
165

Solution:
A→BReactantProduct

Since the reaction follows the first order kinetics and after 70 minutes the concentration of A was found to be half of its initial concentration. That means 70 minutes is the half life time for the reaction. And we know that for a first order reaction half life time is given as.

t12 = 0.693KK=0.693t12=0.69370×60 sec.

=69307×6×10-6 sec-1= 165 ×10-6 sec-1So, x = 165

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (27 Jul Shift 2)
A→B

The above reaction is of zero order. Half life of this reaction is 50 min. The time taken for the concentration of A to reduce to one-fourth of its initial value is min . (Nearest integer)

Options:

The correct answer is:
75

Solution:
The integrated rate equation of zero order reaction is

kt=A0-At

Assume reaction starts with 1 mole A

t1/2=a2k,K=12×50

For 75% completion

A0=a, At=a4

a-a4=kt

t=34ak=34×100a=75 min

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (01 Feb Shift 2)
A→B

The rate constants of the above reaction at 200 K and 300 K are 0.03 min-1 and 0.05 min-1 respectively. The activation energy for the reaction is J (Nearest integer)

(Given : In 10=2.3

R=8.3 J K-1 mol-1

log 5=0.70

log 3=0.48

log2=0.30

Options:

The correct answer is:
2520

Solution:
If K is known at two different temperatures the activation energy can be calculated as:

At temperature 1: ln K1 = -EaRT1 + ln A

At temperature 2: ln K2 = -EaRT2 + ln A

We can subtract one of these equations from the other:

ln K1 - ln K2 = -EaRT1 + ln A - -EaRT2 + ln A

This equation can further simplified to:

lnK1K2 = EaR1T2 - 1T1

ln 0.050.03=Ea2.305×8.314×1200-1300

⇒ 0.22 =Ea2.305×8.314×1600

⇒ Ea = 2.3 × 8.3 × 600 × 0.22

⇒Ea=2519.88 J⇒Ea=2520 J

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (31 Jan Shift 1)
A bacterial infection in an internal wound grows as N't=N0exp⁡(t) , where the time t is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as dNdt=-5N2 . What will be the plot of N0N vs t after 1 hour?

Options:
A
B
C
D

The correct answer is:

Solution:
From 0 to 1 hour, N'=N0et
From 1 hour onwards dNdt=-5N2
So at t=1 hour, N'=eN0
dNdt=-5N2
∫eN0NN-2dN=-5∫1tdt
1N-1eN0=5t-1
N0N-1e=5N0(t-1)
N0N=5N0(t-1)+1e
N0N=5N0t+1e-5N0
Which is following y=mx+C

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A bag ' $A$ ' contains 2 white and 3 red balls and bag ' $B$ ' contains 4 white and 5 red balls. One ball is drawn at random from a randomly chosen bag and is found to be red. The probability that it was drawn from bag ' $B$ ' was

Options:
A $\frac{5}{14}$
B $\frac{5}{16}$
C $\frac{5}{18}$
D $\frac{25}{52}$

The correct answer is:
$\frac{25}{52}$

Solution:
Let $E_1$ be the event that the ball is drawn from bag $A, E_2$ the event that it is drawn from bag $B$ and $E$ that the ball is red. We have to find $P\left(E_2 / E\right)$
Since both the bags are equally likely to be selected, we have $P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$
Also $P\left(E / E_1\right)=3 / 5$ and $P\left(E / E_2\right)=5 / 9$.
Hence by Bay's theorem, we have
$P\left(E_2 / E\right)=\frac{P\left(E_2\right) P\left(E / E_2\right)}{P\left(E_1\right) P\left(E / E_1\right)+P\left(E_2\right) P\left(E / E_2\right)}$
$=\frac{\frac{1}{2} \cdot \frac{5}{9}}{\frac{1}{2} \cdot \frac{3}{5}+\frac{1}{2} \cdot \frac{5}{9}}=\frac{25}{52}$.

Topic: Mathematics, Probability, JEE Main
A bag contains $2 n+1$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $n+1$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then $n$ is equal to

Options:
A $10$
B $11$
C $12$
D $13$

The correct answer is:
$10$

Solution:
The probability that the toss results is a tail $=\frac{(n+1)}{2(2 n+1)}$
$\therefore 1-\frac{(n+1)}{2(2 n+1)}$ is the probability that the toss result is a head.
$$
\begin{array}{lc}
\therefore & 1-\frac{n+1}{2(2 n+1)}=\frac{31}{42} \\
\Rightarrow & \frac{4 n+2-n-1}{4 n+2}=\frac{31}{42} \\
\Rightarrow & \frac{3 n+1}{4 n+2}=\frac{31}{42} \\
\Rightarrow & 126 n+42=124 n+62 \\
\Rightarrow & 2 n=20 \\
\Rightarrow & n=10
\end{array}
$$

Topic: Mathematics, Probability, TS EAMCET, TS EAMCET 2013
A bag contains $2 n+1$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $n+1$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then $n$ is equal to

Options:
A $10$
B $11$
C $12$
D $13$

The correct answer is:
$10$

Solution:
The probability that the toss results is a tail $=\frac{(n+1)}{2(2 n+1)}$
$\therefore 1-\frac{(n+1)}{2(2 n+1)}$ is the probability that the toss result is a head.
$$
\begin{array}{lc}
\therefore & 1-\frac{n+1}{2(2 n+1)}=\frac{31}{42} \\
\Rightarrow & \frac{4 n+2-n-1}{4 n+2}=\frac{31}{42} \\
\Rightarrow & \frac{3 n+1}{4 n+2}=\frac{31}{42} \\
\Rightarrow & 126 n+42=124 n+62 \\
\Rightarrow & 2 n=20 \\
\Rightarrow & n=10
\end{array}
$$

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2013
A bag contains \(2 n\) coins out of which \(n-1\) are unfair with heads on both sides and the remaining are fair. One coin is picked from the bag at random and tossed. If the probability that head falls in the toss is \(\frac{41}{56}\), then the number of unfair coins in the bag is

Options:
A 18
B 15
C 13
D 14

The correct answer is:
13

Solution:
According to given informations,
\(\begin{aligned}
& \frac{(n-1) \times 1+(n+1) \frac{1}{2}}{2 n}=\frac{41}{56} \\
& \Rightarrow \quad \frac{2 n-2+n+1}{4 n}=\frac{41}{56} \\
& \Rightarrow \quad(3 n-1) 14=41 n \\
& \Rightarrow \quad 42 n-14=41 n \\
& \Rightarrow \quad n=14 \\
\end{aligned}\)
\(\therefore\) Number of unfair coins in the bag is \((n-1)=13\). Hence, option (3) is correct.

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)
A bag contains 2 red, 3 white and 5 black balls, a ball is drawn its colour is noted and replaced. If the probability of getting a red ball for the first time at the nth trial is atleast 225, then the greatest value of n is equal to

Options:
A 4
B 5
C 6
D 7

The correct answer is:
5

Solution:
R→represents a red ball is drawn,

PR=210=15
Probability of getting red ball for the first time at nth trial =45n-115

⇒45n-115≥225
⇒45n-1≥25
n-1≤4⇒n≤5

Topic: Mathematics, Probability, JEE Main
A bag contains \( 3 \) red and \( 3 \) green balls and a person draws out \( 3 \) at random. He then drops \( 3 \) blue balls into the bag and
again draws out \( 3 \) at random. The chance that the \( 3 \) later balls being all of different colours is

Options:
A \( 15 \% \)
B \( 20 \% \)
C \( 27 \% \)
D \( 40 \% \)

The correct answer is:
\( 27 \% \)

Solution:
We have a bag containing 3 red and 3 green balls. Now, he draws three balls from the bag. So, possible cases for the event is as follows:

Clearly, from case 3 and 4, we cannot get balls of different colours. Hence, probability is zero.

Now,

probability of drawing 1 red and 2 green balls is

=C13·C23C36=920.

Then, P1=C12·C11·C13C36=620.

And,

probability of drawing 2 red and 1 green ball is

=C13·C23C36=920.

Then, P2=C11·C12·C13C36=620.

Hence, required potability is

=920×620+920×620.

=2920×620=27100 or 27%.

Topic: Mathematics, Probability, JEE Main
A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its color is observed and this ball along with two additional balls of the same color 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:

Options:
A 34
B 310
C 25
D 15

The correct answer is:
25

Solution:
E: Ball drawn from bag is observed to be Red after A or B has occurred.

PE=PE∩A+PE∩B

⇒PE=PAPEA+PBPEB

=410612+610412

=48120

=25

Topic: Mathematics, Probability, JEE Main, JEE Main 2018 (08 Apr)
A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If σ2 is the variance of X, then 100σ2 is equal to

Options:

The correct answer is:
56

Solution:
Given,

X= Number of white ball drawn,

And bag contains 4 white and 6 black balls, total 3 balls are to be drawn,

So, PX=0=C36C310=16

PX=1=C26×C14C310=12

PX=2=C16×C24C310=310

and PX=3=C06×C34C310=130

X	0	1	2	3

PX	16	12	310	130

So, variance =σ2=∑PiXi2-∑PiXi2

⇒σ2=12+1210+310-12+610+1102

⇒σ2=56100

⇒100σ2=56

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A bag contains 5 balls of unknown colours. A ball is drawn at random from it and is found to be red. Then the probability that all the balls in the bag are red, is

Options:
A 12
B 14
C 25
D 13

The correct answer is:
13

Solution:
Let Ri (i = 1, 2, 3, 4, 5) denotes 1, 2, 3, 4 or 5 red balls in the bag. Let R represent the ball drawn is red, then
P(R1)=P(R2)=P(R3)=P(R4)=P(R5)=15
and PRR1=15, PRR2=25, PRR3=35, PRR4=45
PRR5=1
â´ PR5R=P(R5)âPRR5âi=15P(Ri)âPRRi
=11+15+25+35+45=515=13

Topic: Mathematics, Probability, JEE Main
A bag contains 5 white, 4 black and 2 red balls. Balls are drawn one by one without replacement. The probability that the 5th ball is a red ball, is

Options:
A 211
B 411
C 37
D 611

The correct answer is:
211

Solution:
If the 5th ball is a red ball â either the first four ball are of other colour and 5th ball is red or the first four ball consist of 1 red ball, 3 other ball and 5th ball is red ball
Hence, the required probability = C 9 4 C 11 4 Ã 2 7 + C 9 3 Ã C 2 1 C 11 4 Ã 1 7
=9Ã8Ã7Ã611Ã10Ã9Ã8Ã27+9Ã8Ã7Ã2Ã411Ã10Ã9Ã8Ã17
=655+455=1055=211

Topic: Mathematics, Probability, JEE Main
A bag contains 5 white balls, 3 black balls, 4 yellow balls. A ball is drawn from the bag, its colour is noted and put back into the bag with 5 additional balls of the same colour. The process is repeated. The probability that a yellow ball is drawn in the 1st draw given that a white ball is drawn in the 2nd draw is

Options:
A 955
B 417
C 15
D 211

The correct answer is:
417

Solution:
A= A yellow ball is drawn in the 1st draw
B= A white ball is drawn in the 2nd draw
PAB=PAâ©BPB=412Ã517512Ã1017+312Ã517+412Ã517
=417

Topic: Mathematics, Probability, JEE Main
A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is

Options:
A 57
B 27
C 37
D 56

The correct answer is:
57

Solution:
Let the bag contains black ballsB and non-black balls B¯, then we can have the following possibilities:

6B,0B¯, 5B,1B¯, 4B,2B¯, 3B,3B¯,2B,4B¯

All the events are equally likely, i.e., 15.

Required probability

=15×C26C26+15×C25C2615×C26C26+15×C25C26+15×C24C26+15×C23C26+15×C22C26

=2535=57

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (31 Jan Shift 1)
A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is

Options:
A 14
B 1150
C 15
D 950

The correct answer is:
15

Solution:
Given,

Bag have 6 white and 4 black balls

Now also given die is rolled and the number denotes the number of ball drawn from the bag,

So, probability all drawn balls are white will be,

=16C16C110+C26C210+C36C310+C46C410+C56C510+C66C610

=16610+1545+20120+15210+6252+1210

=5042520=15

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

Options:
A 25
B 27
C 17
D 15

The correct answer is:
27

Solution:
Total Possible cases of balls in the bag could be:

2W,6B,3W,5B,4W,4B,5W,3B,6W,2B

The favourable case is to choose 2W and 2B from the bag having 4W,4B

So, the required probability is given by

PE=C24×C24C28C22×C26C28+C23×C25C28+C24×C24C28+C25×C23C28+C26×C22C28

⇒PE=6×615+3×10+36+30+15

⇒PE=36126

⇒PE=27

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (01 Feb Shift 1)
A bag contains 10 white and 3 black balls. Balls are drawn one by one without replacement till all the black balls are drawn. The probability that the procedure of drawing balls will come to an end at the 7th draw is

Options:
A 15286
B 105286
C 35286
D 7286

The correct answer is:
15286

Solution:
The drawing procedure ends at 7th draw, if 3rd black ball comes at 7th draw. Hence, in 1st six draws there will be 2 black balls and 4 white balls.

So, required Probability =C410×C23C613×17=15286

Topic: Mathematics, Probability, JEE Main
A bag contains 21 markers with numbers 1 to 21. A marker is drawn at random and then replaced and then a second marker is drawn. The probability that the first number is odd and the second is even is

Options:
A 1021
B 1121
C 110441
D 100441

The correct answer is:
110441

Solution:
Probability that the first marker is odd =1121
Probability that the second marker is even =1021
The required probability =1121×1021=110441

Topic: Mathematics, Probability, JEE Main
A bag contains 30 tokens numbered serially from 0 to 29. The number of ways of choosing 3 tokens from the bag, such that the sum on them is 30, is

Options:
A 56
B 44
C 75
D 81

The correct answer is:
75

Solution:
Let, a,b,c be the selected numbers; a+b+c=30; a<b<c.
a=0⇒b,c=1,292,28.....14,16→Total 14 ways.
a=1⇒b,c=2,273,26.....14,15→Total 13 ways.
a=2⇒b,c=3,254,24.....13,15→Total 11 ways.
a=3⇒b,c=4,235,22.....13,14→Total 10 ways.
a=4⇒b,c=5,216,20.....12,14→Total 8 ways.
a=5⇒b,c=6,197,18.....12,13→Total 7 ways.
a=6⇒b,c=7,178,16.....11,13→Total 5 ways.
a=7⇒b,c=8,159,14.....11,12→Total 4 ways.
a=8⇒b,c=9,1310,12→Total 2 ways.
a=9⇒b,c=10,11→Total 1 way.
So, the total number of ways =1+2+4+5+7+8+10+11+13+14=75.

Topic: Mathematics, Permutation Combination, JEE Main
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

$\left(\frac{\text { mean of } \mathrm{X}}{\text { standard deviation of } \mathrm{X}}\right)$ is equal to:

Options:
A 4
B $4 \sqrt{3}$
C $3 \sqrt{2}$
D $\frac{4 \sqrt{3}}{3}$

The correct answer is:
$4 \sqrt{3}$

Solution:
$P($ white ball $)=\frac{30}{40}=\frac{3}{4}, Q$ (red ball) $=\frac{10}{40}=\frac{1}{4}, n=16$

$\frac{\text { Mean of } X}{\text { standard deviation of } X}=\frac{n P}{\sqrt{n P Q}}=\frac{\sqrt{n P}}{\sqrt{Q}}$ $=\sqrt{\frac{16 \times \frac{3}{4}}{\frac{1}{4}}}=\sqrt{48}=4 \sqrt{3}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A bag contains 40 tickets numbered from 1 to 40. Two tickets are drawn from the bag without replacement. The probability that the 2nd ticket is a perfect square given that the 1st ticket was a perfect square is

Options:
A 16
B 539
C 320
D 18

The correct answer is:
539

Solution:
A=1st ticket drawn is a perfect square
B=2nd ticket drawn is a perfect square
PA=640; PA∩B=6C240C2=6×540×39
PBA=PB∩APA=6×540×39×406=539

Topic: Mathematics, Probability, JEE Main
A bag contains a mixed lot of red and blue balls. If two balls are drawn at random, the probability of drawing two red balls is five times the probability of drawing two blue balls and the probability of drawing one ball of each colour is six times the probability of drawing two blue balls. Then, the total number of red and blue balls in the bag is

Options:

The correct answer is:
9

Solution:
Let, a= number of red balls.
b= number of blue balls.
p1= probability of drawing two red balls
=aC2a+bC2=aa-1a+ba+b-1
p2= probability of drawings two blue balls
=bC2a+bC2=bb-1a+ba+b-1
p3= probability of drawing one red and one blue ball
=aC1.bC1a+bC2=2aba+ba+b-1
Given that p1=5p2 and p3=6p2
âaa-1=5bb-1 and 2ab=6bb-1
âa=6,b=3â Total number of balls =9

Topic: Mathematics, Probability, JEE Main
A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is p. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colours is q. If p: q=m :n, where m and n are co-prime, then m+n is equal to

Options:

The correct answer is:
14

Solution:
Given,

A bag contains six balls of different colours, two balls are drawn in succession with replacement,

The probability that both the balls are of the same colour will be p=636

Now Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colours is q,

So, probability will be q=C16×C15×C1464=1201296=554

Now the ratio of p & q will be pq=95

Now, on comparing with pq=mn⇒m+n=14

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (30 Jan Shift 2)