datavorous/entrance-exam-dataset huggingface dataset:
There are 4 white and 3 black balls in a box. In another box, there are 3 white and 4 black balls. An unbiased dice is rolled. If it shows a number less than or equal to 3, then a ball is drawn from the second box, otherwise from the first box. If the ball drawn is black then the possibility that the ball was drawn from the first box is

Options:
A $1 / 2$
B $6 / 7$
C $4 / 7$
D $3 / 7$

The correct answer is:
$3 / 7$

Solution:
Box I $\rightarrow 4 \mathrm{~W} ; 3 \mathrm{~B}$
Box II $\rightarrow 3 \mathrm{~W} ; 4 \mathrm{~B}$
Probability for choosing first box $=\frac{3}{6}=\frac{1}{2}$
Probability for choosing second box $=\frac{1}{2}$
$\therefore$ Required probability $=\frac{\frac{1}{2} \times \frac{3}{7}}{\frac{1}{2} \times \frac{3}{7}+\frac{1}{2} \times \frac{4}{7}}$
$=\frac{\frac{3}{14}}{\frac{3}{14}+\frac{4}{14}}=\frac{\frac{3}{14}}{\frac{7}{14}}=\frac{3}{7}$

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 2)
There are 10 points in a plane. No three of these points are in a straight line. What is the total number of straight lines which can be formed by joining the points?

Options:
A 90
B 45
C 40
D 30

The correct answer is:
45

Solution:
A straight line can be formed by joining 2 points. $\therefore$ Total number of straight lines $={ }^{10} \mathrm{C}_{2}$
$=\frac{10 \times 9}{2 \times 1}=45$

Topic: Mathematics, Permutation Combination, NDA, NDA 2019 (Phase 1)
There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:
$$
\begin{array}{|c|c|c|c|c|c|c|}
\hline \text { Marks } & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline \text { Frequency } & x-2 & x & \mathrm{x}^2 & (x+1)^2 & 2 \mathrm{x} & x+1 \\
\hline
\end{array}
$$
where $x$ is a positive integer. Determine the mean and standard deviation of the marks.

Options:

None

Solution:
$$
\begin{aligned}
&\text { We have }\\
&\Sigma f=(x-2)+x+x^2+\left(x^2+2 x+1\right)+2 x+(x+1)=60\\
&\Rightarrow 2 x^2+7 x-60=0 \Rightarrow(x-4)(2 x+15)=0\\
&\Rightarrow x=4, \frac{-15}{2}\\
&\Rightarrow x=4 \quad[\because x \text { can not be }-\mathrm{ve}]
\end{aligned}
$$
So, $\bar{y}=\frac{168}{60}=2.8$
$$
\text { S.D. }=\sqrt{\left(\frac{546}{60}\right)-(2.8)^2}=\sqrt{9.1-7.84}=\sqrt{1.26}=1.12
$$

Topic: Mathematics, Statistics
There are $\mathrm{n}$ socks in a drawer, of which 3 socks are red. If 2 of the socks are chosen randomly and the probability that
both selected socks are red is $\frac{1}{2}$, then what is the value of n?

Options:
A 3
B 4
C 5
D 6

The correct answer is:
4

Solution:
Number of socks $=\mathrm{n}$ Number of red socks $=3$
Given, $\frac{{ }^{3} C_{2}}{{ }^{n} C_{2}}=\frac{1}{2}$
$\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{2}=3 \times 2=6$
$\Rightarrow \mathrm{n}=4$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
There are two opposing views about origin of modern man. According 10 one view Homo erectus in Asia were the ancestors of modern man. A study of variation of DNA however suggested African origin of modern man. What kind of observation on DNA variation could suggest this?

Options:
A greater variation in Asia than in Africa
B greater variation in Africa than in Asia
C similar variation in Africa and Asia
D variation only in Asia and no variation in Africa.

The correct answer is:
greater variation in Asia than in Africa

Solution:
There are two models about origin of modern man (Homo sapiens sapiens)
(i) Multiregional model: According to this view, modern humans evolved in many parts of the world from regional descendants of Homo erectus who dispersed from Africa between 1 and 2 million years ago.
(ii) Monogenesis model: According to this view, African descendents of Homo erectus gave rise to modern humans.
In late 1980s, Rebecca Cann and other geneticists supported this view on the basis of DNA of living humans. They compared the mitochondrial DNA (mtDNA) of a multiathenic sample of more than 100 people representing four continents. The greater the difference between the mtDNAs of two peoples, the longer ago that mtDNAs diverged from a common source. By using bioinformarics, they concluded that the divergence of mtDNA of Africans from common source began just 200, 000 years ago, much to late to represent the dispersal of Homo erectus. Thus, there are greater variation in Asia than in Africa.

Topic: Biology, Evolution, NEET, NEET 2005
There are two opposing views about origin of modern man. According to one view Homo erectus in Asia were the ancestors of modern man. A study of variations of DNA however suggested African origin of modern man. What kind of observation on DNA variation could suggest this?

Options:
A Greater variation in Africa than in Asia
B Variation only in Asia and no variation in Africa
C Greater variation in Asia than in Africa
D Similar variation in Africa and Asia

The correct answer is:
Greater variation in Asia than in Africa

Solution:
There are two models about origin of modern man (Homo sapiens sapiens) i.e.
(i) Multiregional model : According to this view, Modern humans evolved in many parts of the world from regional descendants of Homo erectus who dispersed from Africa between 1 and 2 million years ago.
(ii) Monogenesis model: According to this view only African descendents of Homo erectus gave rise to modern humans. In late 1980s Rebecca Cann and other geneticists supported this view on the basis of DNA of living humans. They compared the mitochondrial DNA (mtDNA) of a multiathenic sample of more than 100 people representing four continents. The greater the difference between the mtDNAs of two peoples, the longer ago that mtDNAs diverged from a common source. By using bioinformatics, they concluded that the divergence of $m t$ DNA of Africans from common source began just 200,000 years ago, much to late to represent the dispersal of Homo erectus. Thus there are greater variation in Asia than in Africa.

Topic: Biology, Evolution, JIPMER, JIPMER 2007
There is $25 \%$ chance that it rains on any particular day. What is the probability that there is at least one rainy day within a period of 7 days?

Options:
A $1-\left(\frac{1}{4}\right)^{7}$
B $\left(\frac{1}{4}\right)^{7}$
C $\left(\frac{3}{4}\right)^{7}$
D $1-\left(\frac{3}{4}\right)^{7}$

The correct answer is:
$1-\left(\frac{3}{4}\right)^{7}$

Solution:
The probability of rain in one day
$=\frac{25}{100}=\frac{1}{4}$
Probability of getting at least one rainy day within a period of 7 days
$=1-\left[1-\frac{1}{4}\right]^{7}=1-\left[\frac{3}{4}\right]^{7}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 2)
There is a forest next to Raju's village which is a natural habitat for an endangered species of lotus. The Government plans to develop a highway through the forest which will cause disturbance in the habitat and may lead to extinction of the endangered species. Being an environmental activist, what must have Raju suggested the Government in order to save the species as the highway project is also important for the development of the village?

Options:
A In situ conservation
B Migration
C Ex situ conservation
D Deforestation

The correct answer is:
Ex situ conservation

Solution:
In ex situ conservation endangered animals and plants that are threatened to become extinct are removed from their natural habitat and allocated in special setting such as zoological parks, botanical gardens and wildlife safari parks where they can be protected. In in situ conservation, the endangered species are protected in their natural habitat so that the entire ecosystem is protected.

Topic: Biology, Biodiversity and its Conservation, NEET
There is a point inside a circle. What is the probability that this point is close to the circumference than to the centre?

Options:
A $3 / 4$
B $1 / 2$
C $1 / 4$
D $1 / 3$

The correct answer is:
$3 / 4$

Solution:
Let radius of circle be ' $r$ '. Total possible outcomes $=$ Area of circle $=\pi r^{2}$ Observe the figure, we have to find the probability of point, $\mathrm{P}$ in the ring which will be closer to circumference.
Area of ring $=$ Area of outer circle $-$ Area of inner circle
$=\pi r^{2}-\pi\left(\frac{r}{2}\right)^{2}=\pi r^{2}-\frac{\pi r^{2}}{4}=\frac{3 \pi r^{2}}{4}$
So, favourable outcome $=\frac{3 \pi r^{2}}{4}$.
Required probability $=\frac{\frac{3}{4} \pi r^{2}}{\pi r^{2}}=\frac{3}{4}$

Topic: Mathematics, Probability, NDA, NDA 2011 (Phase 2)
Thermodynamic standard conditions of temperature and pressure are

Options:
A $0^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$
B $273 \mathrm{~K}$ and $101.3 \mathrm{kPa}$
C $298 \mathrm{~K}$ and $1 \mathrm{~atm}$
D $0^{\circ} \mathrm{C}$ and $101.3 \mathrm{kPa}$

The correct answer is:
$0^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$

Solution:
Thermodynamic standard condition of temperature and pressure is $273 \mathrm{~K}\left(0^{2} \mathrm{C}\right)$ and $1 \mathrm{~atm}(760 \mathrm{~mm} \mathrm{Hg})$.

Topic: Chemistry, Thermodynamics (C), COMEDK, COMEDK 2020
Thevertices of a cubeare $(0,0,0),(2,0,$, 0), $(0,2,0),(0,0,2),(2,2,0),(2,0,2),(0,2,2),(2,2,2)$ respectively.
What is the angle between any two diagonals of the cube?

Options:
A $\cos ^{-1}(1 / 2)$
B $\cos ^{-1}(1 / 3) \quad$
C $\cos ^{-1}(1 / \sqrt{3})$
D $\cos ^{-1}(2 / \sqrt{3})$

The correct answer is:
$\cos ^{-1}(1 / 3) \quad$

Solution:
Required angle $=\cos ^{-1}\left(\frac{1}{3}\right)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2010 (Phase 1)
Thevertices of a cubeare $(0,0,0),(2,0,$, 0), $(0,2,0),(0,0,2),(2,2,0),(2,0,2),(0,2,2),(2,2,2)$ respectively.
What is the angle between one of the edges of the cube and the diagonal of the cube intersecting the edge of the cube?

Options:
A $\cos ^{-1}(1 / 2)$
B $\cos ^{-1}(1 / 3) \quad$
C $\cos ^{-1}(1 / \sqrt{3})$
D $\cos ^{-1}(2 / \sqrt{3})$

The correct answer is:
$\cos ^{-1}(2 / \sqrt{3})$

Solution:
Required angle $=\cos ^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2010 (Phase 1)
Thevertices of a cubeare $(0,0,0),(2,0,$, 0), $(0,2,0),(0,0,2),(2,2,0),(2,0,2),(0,2,2),(2,2,2)$ respectively.
What is the angle between the diagonal of one of the faces of the cube and the diagonal of the cube intersecting the diagonal of the face of the cube?

Options:
A $\cos ^{-1}(1 / \sqrt{3})$
B $\cos ^{-1}(2 / \sqrt{3})$
C $\cos ^{-1}(\sqrt{2 / 3})$
D $\cos ^{-1}(\sqrt{2} / 3)$

The correct answer is:
$\cos ^{-1}(\sqrt{2 / 3})$

Solution:
Required angle $=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2010 (Phase 1)
This flowchart is a simplified model of Phosphorous cycle. Identify labeling A and B.

Options:
A A: Producers
B: Leaching
B A: Leaching
B: Producers
C A: Weathering
B: Producers
D A: Producers
B: Weathering

The correct answer is:
A: Producers
B: Weathering

Solution:
This is a terrestrial ecosystem as it begins with rocks. Phosphorus is present in rocs as phosphates. When they weather, a small amount of phosphates dissolve in soil and reach the plants. They are absorbed by the roots of the plants. Animals will feed on plants and obtain phosphorus. When plants and animals die they are decomposed by bacteria releasing phosphorus.

Topic: Biology, Ecosystem, NEET
Three alcohols
(i) CH3CH2CH2OH,
(ii) CH3-CHOH-CH3 and
(iii) CH3-C(CH3)(OH)-CH3
were treated seperately with Lucas reagent (Conc. HCl+ZnCl2). What results do you expect at room temperature

Options:
A (i) reacts in about 5 minutes, (ii) reacts in about 15 minutes and (iii) not at all
B (iii) reacts immediately, (ii) reacts in about 5 minutes and (i) not at all
C (ii) and (iii) react immediately and (i) in about 5 minutes
D (i) reacts immediately, (ii) reacts in about 5 minutes and (iii) not at all

The correct answer is:
(iii) reacts immediately, (ii) reacts in about 5 minutes and (i) not at all

Solution:
Lucas reagent (Conc. HCl+ZnCl2). is used to distinguish between alcohols tertiary alcohol reacts immediately and gives white turbidity, secondary alcohol reacts in about 5 minutes and primary alcohol do not react.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Three candidates solve a question. Odds in favour of the correct answer are $5: 2,4: 3$ and $3: 4$ respectively for the three candidates. What is the probability that at least two of them solve the question correctly?

Options:
A $209 / 343$
B $134 / 343$
C $149 / 343$
D $60 / 343$

The correct answer is:
$209 / 343$

Solution:
Odd in fav. for student $(\mathrm{A})=\frac{5}{2}=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}\left(\mathrm{A}^{\prime}\right)}$
Odd in fav. for student $(\mathrm{B})=\frac{4}{3}=\frac{\mathrm{P}(\mathrm{B})}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}$
Odd in fav. for student $(\mathrm{C})=\frac{3}{4}=\frac{\mathrm{P}(\mathrm{C})}{\mathrm{P}\left(\mathrm{C}^{\prime}\right)}$
$\Rightarrow \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{2}{5} \mathrm{P}(\mathrm{A}), \mathrm{P}\left(\mathrm{B}^{\prime}\right)=\frac{3}{4} \mathrm{P}(\mathrm{B}), \mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{4}{3} \mathrm{P}(\mathrm{C})$
Now $\mathrm{P}(\mathrm{A})+\mathrm{P}\left(\mathrm{A}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{A})+\frac{2}{5} \mathrm{P}(\mathrm{A})=1 \Rightarrow \mathrm{P}(\mathrm{A})=\frac{5}{7}$
Also $\mathrm{P}(\mathrm{B})+\mathrm{P}\left(\mathrm{B}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{B})+\frac{3}{4} \mathrm{P}(\mathrm{B})=1 \Rightarrow \mathrm{P}(\mathrm{B})=\frac{4}{7}$
And $\mathrm{P}(\mathrm{C})+\mathrm{P}\left(\mathrm{C}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{C})+\frac{4}{3} \mathrm{P}(\mathrm{C})=1 \Rightarrow \mathrm{P}(\mathrm{C})=\frac{3}{7}$
$\therefore \quad \mathrm{P}(\mathrm{A})=\frac{2}{5} \times \frac{5}{7}=\frac{2}{7}, \quad \mathrm{P}\left(\mathrm{B}^{\prime}\right)=\frac{3}{4} \times \frac{4}{7}=\frac{3}{7}$,
$\mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{4}{3} \times \frac{3}{7}=\frac{4}{7}$
Req. Prob. $=\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}\left(\mathrm{C}^{\prime}\right)+\mathrm{P}(\mathrm{A}) \times \mathrm{P}\left(\mathrm{B}^{\prime}\right) \times \mathrm{P}(\mathrm{C})$
$+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{C})$
$=\frac{5}{7} \times \frac{4}{7} \times \frac{4}{7}+\frac{5}{7} \times \frac{3}{7} \times \frac{3}{7}+\frac{2}{7} \times \frac{4}{7} \times \frac{3}{7}+\frac{5}{7} \times \frac{4}{7} \times \frac{3}{7}$
$=\frac{209}{343}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 2)
Three coins are tossed simultaneously. What is the probability that they will fall two heads and one tail?

Options:
A $1 / 3$
B $1 / 2$
C $1 / 4$
D $3 / 8$

The correct answer is:
$3 / 8$

Solution:
Three coins tossed simultaneously $\therefore \quad$ Total outcomes $=2^{3}=8$
Now, S = {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT}
Favourbale cases $=$ two heads one tail $=\mathrm{HHT} . \mathrm{HTH}$. ТНН
$\therefore$ Required prob $=\frac{3}{8}$.

Topic: Mathematics, Probability, NDA, NDA 2012 (Phase 2)
Three dice are thrown simultaneously. What is the probability that the sum on the three faces is at least $5 ?$

Options:
A $\frac{17}{18}$
B $\frac{53}{54}$
C $\frac{103}{108}$
D $\frac{215}{216}$

The correct answer is:
$\frac{53}{54}$

Solution:
As we know that 3 dice are thrown. We want prob. of sum on three faces at least 5 i.e. some may be 5 or more. We will find prob. of sum on three faces not 5 or less.
i.e. sum on faces is 3 and $4(1,2$ is not possible because of 3 dice). No. of ways for sum on faces not 5 or more $=4$ $[(1,1,1),(1,2,1),(1,1,2),(2,1,1)]$
Total out comes $=216$
Prob. of not 5 or more $=\frac{4}{216}$ Prob. of sum on three faces at least 5
$=1-\frac{4}{216}=\frac{212}{216}=\frac{53}{54}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 1)
Three dice are thrown. What is the probability of getting a triplet ?

Options:
A $\frac{1}{6}$
B $\frac{1}{18}$
C $\frac{1}{36}$
D $\frac{1}{72}$

The correct answer is:
$\frac{1}{36}$

Solution:
Number of possible outcomes $=216$ triplets $=(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5)$,
$(6,6,6)$
Required probability $=\frac{6}{216}=\frac{1}{36}$

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 2)
Three dice are thrown. What is the probability that the same number will appear on each of them?

Options:
A $\frac{1}{6}$
B $\frac{1}{18}$
C $\frac{1}{24}$
D $\frac{1}{36}$

The correct answer is:
$\frac{1}{36}$

Solution:
Total no. of case $=6^{3}=216$ Favourable cases $=\{(1,1,1),(2,2,2),(3,3,3),(4,4,4),$,
$(5,5,5),(6,6,6)\}$
Probability $=\frac{6}{216}=\frac{1}{36}$

Topic: Mathematics, Probability, NDA, NDA 2012 (Phase 1)
Three dice having digits $1,2,3,4,5$ and 6 on their faces are marked I, II and III and rolled. Let $x, y$ and $z$ represent the number on die-I die-II and die -III respectively. What is the number of possible outcomes such that $x>y>z ?$

Options:
A 14
B 16
C 18
D $\underline{20}$

The correct answer is:
$\underline{20}$

Solution:

Topic: Mathematics, Permutation Combination, NDA, NDA 2018 (Phase 2)
Three digital numbers are formed using the digits $0,2,4,6,8$. A number is chosen at random out of these numbers. What is the probability that the number has the same digits?

Options:
A $1 / 16$
B $1 / 25$
C $16 / 25$
D $1 / 645$

The correct answer is:
$1 / 25$

Solution:
Given digits are $0,2,4,6,8$. Total number of cases $=4 \times 4 \times 4$
(\becauseWe have 4 choices for each number)
Required probability $=\frac{4}{64}=\frac{1}{16}$

Topic: Mathematics, Probability, NDA, NDA 2010 (Phase 2)
Three digits are chosen at random from $1,2,3,4,5,6,7,8$ and 9 without repeating any digit. What is the probability that the product is odd ?

Options:
A $\frac{2}{3}$
B $\frac{7}{48}$
C $\frac{5}{42}$
D $\frac{5}{108}$

The correct answer is:
$\frac{5}{42}$

Solution:
Total no. of 3 -digit numbers $=9 \times 8 \times 7=504$ For product to be odd, we have to choose only from odd numbers. $\therefore$ Total no. of 3 -digit no. whose product are odd $=5 \times 4 \times 3=60$
$\therefore$ Required probability $=\frac{60}{504}=\frac{5}{42}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 2)
Three expressions are given below:
$Q_{1}=\sin (A+B)+\sin (B+C)+\sin (C+A)$
$Q_{2}=\cos (A-B)+\cos (B-C)+\cos (C-A)$
$Q_{3}=\sin A(\cos B+\cos C)+\sin B(\cos C+\cos A)+$
$\sin C(\cos A+\cos B)$
Which one of the following is correct?

Options:
A $Q_{1}=Q_{2}$
B $Q_{2}=Q_{3}$
C $Q_{1}=Q_{3}$
D All the expressions are different

The correct answer is:
$Q_{1}=Q_{3}$

Solution:
We take \mathrm{Q}_{3} \text { first, }\end{array}$
$\mathrm{Q}_{3}=\sin \mathrm{A}(\cos \mathrm{B}+\cos \mathrm{C})+\sin \mathrm{B}(\cos \mathrm{C}+\cos \mathrm{A})+\sin$
$\mathrm{C}(\cos \mathrm{A}+\cos \mathrm{B})$
$=\sin \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \cos \mathrm{C}+\sin \mathrm{B} \cos \mathrm{C}+\sin \mathrm{B} \cos \mathrm{A}$
$+\sin \mathrm{C} \cos \mathrm{A}+\sin \mathrm{C} \cos \mathrm{B}$
$=\operatorname{Qin}(\mathrm{A}+\mathrm{B})+\sin (\mathrm{B}+\mathrm{C})+\sin (\mathrm{C}+\mathrm{A})=\mathrm{Q}_{1}$
$\Rightarrow \mathrm{Q}_{3}=\mathrm{Q}_{1}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2006 (Phase 1)
Three groups of children contain 3 girls and 1 boy; 2 girls and 2 boys; 1 girl and 3 boys. One child is selected at random from each group. The probability that the three selected consist of 1 girl and 2 boys is

Options:
A $\frac{13}{32}$
B $\frac{9}{32}$
C $\frac{3}{32}$
D $\frac{1}{32}$

The correct answer is:
$\frac{13}{32}$

Solution:
Required probability
$=\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4}+\frac{1}{4} \times \frac{2}{4} \times \frac{3}{4}+\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4}$
$=\frac{26}{64}=\frac{13}{32}$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 2)
Three independent events, $\mathrm{A}_{1}, \mathrm{~A}_{2}$ and $\mathrm{A}_{3}$ occur with probabilities $\mathrm{P}\left(\mathrm{A}_{\mathrm{i}}\right)=\frac{1}{1+\mathrm{i}}, \mathrm{i}=1,2,3$. What is the probability
that at least one of the three events occurs?

Options:
A $\frac{1}{4}$
B $\frac{2}{3}$
C $\frac{3}{4}$
D $\frac{1}{24}$

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

Solution:
$P\left(A_{1}\right)=\frac{1}{1+1}=\frac{1}{2}$
$P\left(A_{2}\right)=\frac{1}{3}$
$P\left(A_{3}\right)=\frac{1}{4}$
$\therefore \quad$ Probability that at least one of these events occur is $P\left(A_{1} \cup A_{2} \cup A_{3}\right)$. Also $A_{1}, A_{2} \& A_{3}$ are independent
events.
$\begin{aligned} P\left(A_{1} \cup A_{2} \cup A_{3}\right)=& P\left(A_{1}\right)+P\left(A_{2}\right)+P\left(A_{3}\right) \\ &-P\left(A_{1} \cap A_{2}\right)-P\left(A_{1} \cap A_{3}\right) \\ &-P\left(A_{2} \cap A_{3}\right)+P\left(A_{1} \cap A_{2} \cap A_{3}\right) \\=\frac{1}{2} &+\frac{1}{3}+\frac{1}{4}-\left(\frac{1}{2} \times \frac{1}{3}\right)-\left(\frac{1}{2} \times \frac{1}{4}\right) \\ &-\left(\frac{1}{3} \times \frac{1}{4}\right)+\left(\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}\right) \\=& \frac{3}{4} . \end{aligned}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 2)
Three letters are randomly selected fromthe 26 capital letters of the English Alphabet. What is the probability that the letter 'A'will not be included in the choice?

Options:
A $1 / 2$
B $23/26$
C $12 / 13$
D $25 / 26$

The correct answer is:
$23/26$

Solution:
Total no. of letters $=26$
No. of selected letters $=3$ $\therefore$ No. of ways to select 3 letters out of 26 letters
$={ }^{26} C_{3}$
Since, A will not be include in our choice therefore, Total no. of letters $=26-1=25$ Now, No. of ways to select 3 letters out of 25 letters
${ }^{25} C_{3}$
$\therefore$ Required Prob $=\frac{{ }^{25} C_{3}}{{ }^{6} C_{3}}=\frac{25 !}{3 ! 22 !} \times \frac{3 ! 23 !}{26 !}=\frac{23}{26}$

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 2)
Three resistances $P, Q, R$ each of $2 \Omega$ and an unknown resistance $S$ form the four arms of a Wheatstone's bridge circuit. When a resistance of $6 \Omega$ is connected in parallel to $S$ the bridge gets balanced. What is the value of $S$ ?

Options:
A $2 \Omega$
B $3 \Omega$
C $6 \Omega$
D $1 \Omega$

The correct answer is:
$3 \Omega$

Solution:
The situation can be depicted as shown in figure.
As resistances $S$ and $6 \Omega$ are in parallel their effective resistance is
$\frac{6 S}{6+S} \Omega$
As the bridge is balanced, hence it is balanced Wheatstone's bridge.
For balancing condition,
$\begin{aligned}
& & \frac{P}{Q} & =\frac{R}{\left(\frac{6 S}{6+S}\right)} \\
\text { or } & & \frac{2}{2} & =\frac{2(6+S)}{6 S} \\
\text { or } & & 3 S & =6+S \\
\text { or } & & S & =3 \Omega
\end{aligned}$

Topic: Physics, Current Electricity, JIPMER, JIPMER 2009
Three sides of a trapezium are each equal to $6 \mathrm{~cm}$. Let $\alpha \in\left(0, \frac{\pi}{2}\right)$ be the angle between a pair of adjacent sides.
If the area of the trapezium is maximum, what is the length of the fourth side?

Options:
A $8 \mathrm{~cm}$
B $9 \mathrm{~cm}$
C $10 \mathrm{~cm}$
D $12 \mathrm{~cm}$

The correct answer is:
$12 \mathrm{~cm}$

Solution:
Fourth side, $\mathrm{DC}=\mathrm{x}+6+\mathrm{x}$
$=12 .$

Topic: Mathematics, Area Under Curves, NDA, NDA 2018 (Phase 1)