Probability Practice

1) Out of 2121 tickets consecutively numbered , three are drawn at random . Find the probability that the numbers on them are in A.P.

2) If 22 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 22 points is greater than L3\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+nm + n times (m>n)(m >n) . What is the probability that at least m consecutive heads come up.

5) From an urn containing six balls, 33 white and 33 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) 55 different marbles are placed in 55 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 44 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 22 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 AA and BB be 22 independent witnesses in a case . The probability that AA will speak the truth is x an the probability that BB will speak the truth is y. AA and BB agree in a certain statement. Probability of true statement-

10) A artillery target may be either at a point AA with probability 89\dfrac{8}{9} or at a point BB with probability 19\dfrac{1}{9}. We have 21 shells each of which can be fixed either at point AA or BB . Each shell may hit the target independently of the other with probability 12\dfrac{1}{2}. How many shells must be fixed at point A to hit the target with maximum probability.

11) If aa and bb are chosen randomly from the set consisting of numbers 1,2,3,4,5,61 , 2, 3, 4, 5, 6 with replacement . Find the probability that limx0(ax+bx2)2x=6 \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,,3n)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 P1P_{1} and P2P_{2} are playing the fina of a chess championship , which consisits of a series of matches . Probability of P1P_{1} winning a match is 23\dfrac{2}{3} and for P2P_{2} is 13\dfrac{1}{3} . The winner will be the one who is ahead by 2 games as compared to the other player and wins atleast 66 games . Now , if the player P2P_{2} wins first four matches , find the probability of P1P_{1} winning the championship.

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


15) A point is selected at random from the interior of the pentagon with vertices A=(0,2),B=(4,0),C=(2π+1,0),D=(2π+1,4),E=(0,4)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 APBAPB is obtuse .


16) There are two bags , each containing 55 red and 33 black balls . Two persons AA and BB 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.n.

17) Each of n urns contains aa white and bb 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 , 2n2^n equally skilled players namely S1,S2,S3,,S2nS_{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 S2S_{2} reaches semi - final , then find the probability that S1S_{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 AA and BB play tournament. The first one to win (n+1)(n+1) games , win the series . The probability that AA wins a game is pp and that BB wins a game is qq ( no ties) . Find the probability that A wins the series.


Note by U Z
4 years, 8 months ago

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1 vote

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@megh choksi Nice set of questions. Remember having solved a lot of these during my JEE days.

Sudeep Salgia - 4 years, 8 months ago

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Sir which question you founded the most interesting?

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

U Z - 4 years, 8 months ago

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Hi @megh choksi I found something that might be of some interest to you , Look this up !!

A Former Brilliant Member - 4 years, 8 months ago

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

A Former Brilliant Member - 4 years, 8 months ago

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Nope. They are in the last week of February.

Sudeep Salgia - 4 years, 8 months ago

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@Sudeep Salgia Wish you the Best of Luck !!

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

A Former Brilliant Member - 4 years, 8 months ago

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Q 17

Is the answer aa+b\frac{a}{a+b} ?

Proof :

We are interested in finding the probability of getting a black ball in the last picking up, Let Pk1P_{k-1} be the probability that a Black Ball get's drawn in the (k1)th(k-1)^{th} drawing .(Here we may denote Qk1Q_{k-1} as the probability of it's negation and use it but it turns out after solving , it would be better if leave Qk1Q_{k-1} as 1Pk11-P_{k-1}) .

Pk=(Pk1)(a+1)a+b+1+a(1Pk1)a+b+1=Pk1a+b+1+aa+b+1=P1(a+b+1)k1+i=2k1a(a+b+1)ki=aa+b\begin{aligned} 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{aligned}

A Former Brilliant Member - 4 years, 8 months ago

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Perfect! Looking for proof!

Pranjal Jain - 4 years, 8 months ago

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Check out page 12 over here.

Pratik Shastri - 4 years, 8 months ago

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Q 1

There is a general formula for this question especially for drawing 3 tickets from 2n+12n+1 total tickets .

3n4n21\dfrac{3n}{4n^{2} - 1 }

2n+1=212n+1=21 n=10 n=10

The answer is P=10133 P=\dfrac{10}{133} .

A Former Brilliant Member - 4 years, 8 months ago

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Q 2) The answer I get is 79\dfrac{7}{9}. For Q 3), I get n2n1\dfrac{n}{2^{n-1}}.

Pratik Shastri - 4 years, 8 months ago

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Perfect

U Z - 4 years, 8 months ago

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Yes, even I'm getting the same answer for Q 3

A Former Brilliant Member - 4 years, 8 months ago

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That's correct.

U Z - 4 years, 8 months ago

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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)=(qp)n×(P(1)1)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

A Former Brilliant Member - 4 years, 8 months ago

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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 5π28π=516\frac{\frac{5\pi}{2}}{8\pi} =\frac{5}{16} .

A Former Brilliant Member - 4 years, 8 months ago

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That's the way I guess.

Pratik Shastri - 4 years, 8 months ago

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Q 7

Using Gap method ,the Blue counters can go in 6 places, therefore (64)=15\binom{6}{4}=15 .

Total ways to arrange 9 counters (94)=126\binom{9}{4}=126

ANS=542\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!3! 2! \cdot 3! but NOTE you'll also have to divide by 2!3!2! \cdot 3!

A Former Brilliant Member - 4 years, 8 months ago

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I am getting total ways=150

Pranjal Jain - 4 years, 8 months ago

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Would you please post a complete solution?

A Former Brilliant Member - 4 years, 8 months ago

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@A Former Brilliant Member Lets arrange 2 red and 3 green first. Number of ways=5!2!3!=10\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, (64)=15\binom{6}{4}=15. So the answer must be 10×15=150

Pranjal Jain - 4 years, 8 months ago

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@Pranjal Jain 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 by @megh choksi

A Former Brilliant Member - 4 years, 8 months ago

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@Pranjal Jain True. And the total ways of arrangement are 9!2!.3!.4!=1260\displaystyle \frac{9!}{2!.3!.4!} = 1260 . So when you can calculate the probability it would come out to be the same as calculated by Azhaghu.

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 - 4 years, 8 months ago

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@Sudeep Salgia Thanks for clarifying .

A Former Brilliant Member - 4 years, 8 months ago

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You also need to arrange the red and green counters..

Pratik Shastri - 4 years, 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!×3! 2! \times 3! . Your method is correct.

Siddhartha Srivastava - 4 years, 8 months ago

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Could you please elaborate how you got the total number of ways??

Sudeep Salgia - 4 years, 8 months ago

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Q 7 (ALITER)

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_ \_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 , (64)4!\binom{6}{4} 4! .

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

Final answer = (64)4!×5!9!=15126=542\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 !!!

A Former Brilliant Member - 4 years, 8 months ago

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Q 10

The answer 12 shells .

Proof coming up soon !!

A Former Brilliant Member - 4 years, 8 months ago

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yes , it was very interesting question

I will upvote your comment if your answer is correct as you know I can't comment to every comment

U Z - 4 years, 8 months ago

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Ok

A Former Brilliant Member - 4 years, 8 months ago

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Q 4

Is the answer n+22m+2 \dfrac{n+2}{2^{m+2}} ?

A Former Brilliant Member - 4 years, 8 months ago

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Q 9

The answer is xy1xy+2xy\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 .

A Former Brilliant Member - 4 years, 8 months ago

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It pretty simple.

Probability that AA and BB agree = Probability that both say it is false + Probability that both say it is true

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

And from above we can also say that the probability of both agreeing it to be true = xyxy .

Hence the required probability = xy(1x)(1y)+xy=xy1xy+2xy\displaystyle \frac{xy}{(1-x)(1-y) + xy } = \frac{xy}{1-x -y + 2xy } .

Sudeep Salgia - 4 years, 8 months ago

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Yeah, it is indeed simple !! Thanks

A Former Brilliant Member - 4 years, 8 months ago

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