1) Out of \(21\) tickets consecutively numbered , three are drawn at random . Find the probability that the numbers on them are in A.P.

2) If \(2\) points are selected on a line of length L so as to be on opposite sides of the midpoint of the line , find the probability that the distance between the \(2\) points is greater than \(\dfrac{L}{3}\).

3) If n points are independently chosen at random on the circumference of a circle , what is the probability that these points lie in some semicircle.

4) A coin is tossed \(m + n\) times \((m >n)\) . What is the probability that at least m consecutive heads come up.

5) From an urn containing six balls, \(3\) white and \(3\) black ones, a p erson selects at random an even number of balls (all the different ways of drawing an even number of balls are considered equally probable, irrespective of their number). Then the probability that there will be the same number of black and white balls among them.

6) \(5\) different marbles are placed in \(5\) different boxes randomly. Find the probability that exactly two boxes remain empty. (each box can hold any number of marbles)

7) Two red counters, three green counters and \(4\) blue counters are placed in a row in random order. The probability that no two blue counters are adjacent is

8) Find the probability of ruin of each of \(2\) players when they continue a certain game till the ultimate ruin of on of them.The winner of each game gets one ruble.

9) Let \(A\) and \(B\) be \(2\) independent witnesses in a case . The probability that \(A \)will speak the truth is x an the probability that \(B\) will speak the truth is y. \(A\) and \(B\) agree in a certain statement. Probability of true statement-

10) A artillery target may be either at a point \(A\) with probability \(\dfrac{8}{9}\) or at a point \(B\) with probability \(\dfrac{1}{9}\). We have 21 shells each of which can be fixed either at point \(A\) or \(B\) . Each shell may hit the target independently of the other with probability \(\dfrac{1}{2}\). How many shells must be fixed at point A to hit the target with maximum probability.

11) If \(a\) and \(b\) are chosen randomly from the set consisting of numbers \(1 , 2, 3, 4, 5, 6\) with replacement . Find the probability that \( \displaystyle lim_{x \to 0} \left( \dfrac{a^x + b^x}{2} \right)^{\dfrac{2}{x}} = 6\)

12) From the set \(S = ( 1 , 2 , 3 , \cdots , 3n ) \) , three numbers are chosen at random . Find the probability that the sum of the chosen numbers is divisible by 3.

13) Two players \(P_{1}\) and \(P_{2}\) are playing the fina of a chess championship , which consisits of a series of matches . Probability of \(P_{1}\) winning a match is \(\dfrac{2}{3}\) and for \(P_{2}\) is \(\dfrac{1}{3}\) . The winner will be the one who is ahead by 2 games as compared to the other player and wins atleast \(6\) games . Now , if the player \(P_{2}\) wins first four matches , find the probability of \(P_{1}\) winning the championship.

14) Let \(X\) be a set containing n elements . Find the number of all ordered triplets \((A, B , C)\) of subsets of \(X\) such that \(A\) is a subset of \(B\) and \(B\) is a proper subset of \(C\).

**15) A point is selected at random from the interior of the pentagon with vertices \(A = (0,2) , B = (4 , 0) , C = (2 \pi + 1 , 0) , D = ( 2 \pi + 1 , 4) , E = (0 , 4)\) . What is the probability that angle \(APB\) is obtuse .**

16) There are two bags , each containing \(5\) red and \(3\) black balls . Two persons \(A\) and \(B\) are given one bag each . Each of them is to draw one ball at random from the bag till both of them get a black ball ( not necessary in the same draw). The balls are to be replaced after each draw . Find the probability that the number of trials required is \(n.\)

17) Each of n urns contains \(a\) white and \(b\) blac balls . One ball is transferred from the first urn into the second , then one ball from the latter into the third and so on . If finally one ball is taken out from the last urn , then what is the probability of its being black.

18) In a knockout tournament , \(2^n\) equally skilled players namely \(S_{1} , S_{2} , S_{3} , \cdots , S_{2^n}\) are participating. In each round , players are divided in pairs at random and winner from each pair moves in the next round . If \(S_{2}\) reaches semi - final , then find the probability that \(S_{1}\) will win the tournament.

19) If a pair of dice is thrown twice , then what is the probability that the sum of outcomes of each of the two throws are equal.

20) Two teams \(A\) and \(B\) play tournament. The first one to win \((n+1)\) games , win the series . The probability that \(A\) wins a game is \(p\) and that \(B\) wins a game is \(q\) ( no ties) . Find the probability that A wins the series.

## Comments

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TopNewest@megh choksi Nice set of questions. Remember having solved a lot of these during my JEE days. – Sudeep Salgia · 1 year, 8 months ago

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What was your favorite concepts in math so far?

Can you make a note on a topic which you founded the most beautiful and enjoyable in your own words?

Thanks – Megh Choksi · 1 year, 8 months ago

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@megh choksi I found something that might be of some interest to you , Look this up !! – Azhaghu Roopesh M · 1 year, 8 months ago

HiLog in to reply

– Azhaghu Roopesh M · 1 year, 8 months ago

Hi, are your mid-semester exams over ? If yes, how did you fare in them ?Log in to reply

– Sudeep Salgia · 1 year, 8 months ago

Nope. They are in the last week of February.Log in to reply

And I have a favour to ask of you . I had recently asked Preshit Wazalwar to join Brilliant and I even said that you were here on Brilliant so it would be great if you tell him what he is missing out on and ask him to join Brilliant as soon as possible(well, after the exams ofc) !!

Thanks – Azhaghu Roopesh M · 1 year, 8 months ago

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Q 17

Is the answer \(\frac{a}{a+b}\) ?

Proof :

We are interested in finding the probability of getting a black ball in the last picking up, Let \(P_{k-1}\) be the probability that a Black Ball get's drawn in the \((k-1)^{th}\) drawing .(Here we may denote \(Q_{k-1}\) as the probability of it's negation and use it but it turns out after solving , it would be better if leave \(Q_{k-1}\) as \(1-P_{k-1}\)) .

\[\begin{align} P_k = \dfrac{(P_{k-1})(a+1)}{a+b+1} + \dfrac{a(1-P_{k-1})}{a+b+1} \\& = \dfrac{P_{k-1}}{a+b+1} + \dfrac{a}{a+b+1} \\& = \dfrac{P_1}{(a+b+1)^{k-1}} + \sum_{i=2}^{k-1} \dfrac{a}{(a+b+1)^{k-i}} \\& = \dfrac{a}{a+b} \end{align}\] – Azhaghu Roopesh M · 1 year, 8 months ago

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– Pranjal Jain · 1 year, 8 months ago

Perfect! Looking for proof!Log in to reply

over here. – Pratik Shastri · 1 year, 8 months ago

Check out page 12Log in to reply

Q 4

Is the answer \( \dfrac{n+2}{2^{m+2}} \) ? – Azhaghu Roopesh M · 1 year, 8 months ago

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Q 10

The answer 12 shells .

Proof coming up soon !! – Azhaghu Roopesh M · 1 year, 8 months ago

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I will upvote your comment if your answer is correct as you know I can't comment to every comment – Megh Choksi · 1 year, 8 months ago

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– Azhaghu Roopesh M · 1 year, 8 months ago

OkLog in to reply

Q 7

Using Gap method ,the Blue counters can go in 6 places, therefore \(\binom{6}{4}=15\) .

Total ways to arrange 9 counters \(\binom{9}{4}=126\)

ANS=\(\boxed{\frac{5}{42}}\)

** I think that @Pratik Shastri , @Siddhartha Srivastava you guys are confused about arranging the red and green counters by multiplying by \( 2! \cdot 3!\) but NOTE you'll also have to divide by \(2! \cdot 3!\) – Azhaghu Roopesh M · 1 year, 8 months ago

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– Pranjal Jain · 1 year, 8 months ago

I am getting total ways=150Log in to reply

– Azhaghu Roopesh M · 1 year, 8 months ago

Would you please post a complete solution?Log in to reply

– Pranjal Jain · 1 year, 8 months ago

Lets arrange 2 red and 3 green first. Number of ways=\(\dfrac{5!}{2!3!}=10\). Now, we have six places to put blue ones (2 ends and 4 between others) Select 4 out of 6, \(\binom{6}{4}=15\). So the answer must be 10×15=150Log in to reply

So basically you are multiplying by the ways of arranging red and green counters as well and then later dividing by the same number while counting the total number of ways. – Sudeep Salgia · 1 year, 8 months ago

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– Azhaghu Roopesh M · 1 year, 8 months ago

Thanks for clarifying .Log in to reply

@megh choksi – Azhaghu Roopesh M · 1 year, 8 months ago

Thanks but I am still not convinced that we need to arrange the red and green counters since what difference do they make in contributing to separate the blue ones , but I'll wait till an official answer is posted byLog in to reply

– Sudeep Salgia · 1 year, 8 months ago

Could you please elaborate how you got the total number of ways??Log in to reply

Ok, the simplest method of all. I'll still be using the Gap Method though.

ASSUMPTIONS:Let all the counters be DIFFERENT . So total ways of arrangement is 9! .

KEY: 'O' -> Other counters (other than blue) ----- So we have 5 'Other' counters

\[ _O _ O _ O _ O _ O _ \]

We'll be arranging the blue counters in the 6 gaps which have been divided by the 5 O's . We select and then arrange , \(\binom{6}{4} 4!\) .

Now, we also have to arrange the 5 O's , so multiply by another 5! .

Final answer = \(\dfrac{\binom{6}{4}4! \times 5!}{9!} =\frac{15}{126}=\frac{5}{42}\) .

I hope that it is clear now .

In my other solution I assumed counters of the same colour to be similar BUT interestingly the answer comes out to be the same . It would be nice if you could explain it to me how both the answers are same .

Thanks for the same !!! – Azhaghu Roopesh M · 1 year, 8 months ago

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@Azhaghu Roopesh M True. I hadn't checked the calculations and had only seen that you hadn't multiplie by \( 2! \times 3! \). Your method is correct. – Siddhartha Srivastava · 1 year, 8 months ago

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– Pratik Shastri · 1 year, 8 months ago

You also need to arrange the red and green counters..Log in to reply

Q 15

My approach might sound foolish but it is the only one that gave me a proper answer as compared to the other methods that I came up with .

Did you guys @megh choksi , @Pratik Shastri think that it is possible that a circle with AB as diameter can be constructed ? I reckon P must lie inside this circle.

Its probability comes out to be \[\frac{\frac{5\pi}{2}}{8\pi} =\frac{5}{16}\] . – Azhaghu Roopesh M · 1 year, 8 months ago

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– Pratik Shastri · 1 year, 8 months ago

That's the way I guess.Log in to reply

Q 8

I have assumed that P(n) = probability of ruin of A and Q(n)= that of Q . Let n= no. of marbles for A and B each .p=Probability of A winning and q= that for Q .

I'm getting \[P(n+1) -P(n)=(\dfrac{q}{p})^{n}\times (P(1)-1)\]

On further solving I get a very UGLY answer , can you please verify this , Megh ?

@megh choksi – Azhaghu Roopesh M · 1 year, 8 months ago

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Q 2) The answer I get is \(\dfrac{7}{9}\). For Q 3), I get \(\dfrac{n}{2^{n-1}}\). – Pratik Shastri · 1 year, 8 months ago

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– Azhaghu Roopesh M · 1 year, 8 months ago

Yes, even I'm getting the same answer for Q 3Log in to reply

– Megh Choksi · 1 year, 8 months ago

That's correct.Log in to reply

– Megh Choksi · 1 year, 8 months ago

PerfectLog in to reply

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– Pratik Shastri · 1 year, 8 months ago

Ooops I calculated the probability of compliment of the stated event :PLog in to reply

– Azhaghu Roopesh M · 1 year, 8 months ago

Can you post the solution for Q 2?Log in to reply

Now, the probability that the point in the right half of the stick lies in a region such that the mutual distance is less than \(\dfrac{L}{3}\) is \(\dfrac{x-L/6}{L/2}\)

Therefore, \(P'(\text{compliment of the required probability})=\displaystyle\int_{L/6}^{L/2}\dfrac{\mathrm{d}x}{L/2}\dfrac{x-L/6}{L/2}=\dfrac{2}{9}\)

Therefore, the required probability is \(\boxed{\dfrac{2}{9}}\). I beleive this can also be found graphically. – Pratik Shastri · 1 year, 8 months ago

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– Arpit Agarwal · 1 year, 8 months ago

Hi dude, how can you find this result graphically ?Log in to reply

– Azhaghu Roopesh M · 1 year, 8 months ago

Are you the JEE topper from a few years ago?Log in to reply

– Megh Choksi · 1 year, 8 months ago

His location is jaipur , and the one you are talking about lives in delhiLog in to reply

– Azhaghu Roopesh M · 1 year, 8 months ago

Oh I see . And hey, check out the new problem that I posted .Log in to reply

– Azhaghu Roopesh M · 1 year, 8 months ago

Oh , ok. Thanks.Log in to reply

– Azhaghu Roopesh M · 1 year, 8 months ago

Hi Megh, are some of these questions taken from somewhere ? Just a thoughtLog in to reply

– Megh Choksi · 1 year, 8 months ago

Yes all are taken from my monthly magazine - Mathematics Today Concepts Boosters Alok KumarLog in to reply

– Azhaghu Roopesh M · 1 year, 8 months ago

Yeah, I was really surprised that you had thought up all these good questions in a short span of time ?Log in to reply

– Megh Choksi · 1 year, 8 months ago

It's very rare that I think a question of probabilityLog in to reply

Q 1

There is a general formula for this question especially for drawing 3 tickets from \(2n+1\) total tickets .

\(\dfrac{3n}{4n^{2} - 1 }\)

\[2n+1=21 \] \[ n=10 \]

The answer is \( P=\dfrac{10}{133}\) . – Azhaghu Roopesh M · 1 year, 8 months ago

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Q 9

The answer is \[\dfrac{x \cdot y}{1-x-y+2xy }\] .

But IDK the solution but know the answer since I had got it wrong in a CPT . – Azhaghu Roopesh M · 1 year, 8 months ago

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Probability that \(A\) and \(B\) agree = Probability that both say it is false + Probability that both say it is true

\(\displaystyle = (1-x)(1-y) + xy \).

And from above we can also say that the probability of both agreeing it to be true = \(xy \).

Hence the required probability = \(\displaystyle \frac{xy}{(1-x)(1-y) + xy } = \frac{xy}{1-x -y + 2xy } \). – Sudeep Salgia · 1 year, 8 months ago

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– Azhaghu Roopesh M · 1 year, 8 months ago

Yeah, it is indeed simple !! ThanksLog in to reply