# Probability Practice

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.

Note by U Z
5 years ago

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## Comments

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

- 5 years 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

- 5 years ago

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

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

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Nope. They are in the last week of February.

- 5 years ago

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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

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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{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}

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Perfect! Looking for proof!

- 5 years ago

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Check out page 12 over here.

- 5 years ago

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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 $P=\dfrac{10}{133}$ .

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

- 5 years ago

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Perfect

- 5 years ago

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

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That's correct.

- 5 years 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 $P(n+1) -P(n)=(\dfrac{q}{p})^{n}\times (P(1)-1)$

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

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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 $\frac{\frac{5\pi}{2}}{8\pi} =\frac{5}{16}$ .

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That's the way I guess.

- 5 years ago

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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 $2! \cdot 3!$

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I am getting total ways=150

- 5 years ago

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Would you please post a complete solution?

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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=150

- 5 years ago

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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

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True. And the total ways of arrangement are $\displaystyle \frac{9!}{2!.3!.4!} = 1260$. 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.

- 5 years ago

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Thanks for clarifying .

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

- 5 years 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.

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

- 5 years 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

$\_O \_ O \_ O \_ O \_ O \_$

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 !!!

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Q 10

The answer 12 shells .

Proof coming up soon !!

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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

- 5 years ago

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Ok

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Q 4

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

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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 .

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It pretty simple.

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 = $\displaystyle \frac{xy}{(1-x)(1-y) + xy } = \frac{xy}{1-x -y + 2xy }$.

- 5 years ago

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Yeah, it is indeed simple !! Thanks

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