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How to roll a Dice by tossing a Coin ? Cheenta Statistics Department

How can you roll a dice by tossing a coin? Can you use your probability knowledge? Use your conditioning skills.

Suppose, you have gone to a picnic with your friends. You have planned to play the physical version of the Snake and Ladder game. You found out that you have lost your dice.

The shit just became real!

Now, you have an unbiased coin in your wallet / purse. You know Probability.

Aapna Time Aayega

starts playing in the background. :p

Can you simulate the dice from the coin?

Ofcourse, you know chances better than others. :3

Take a coin.

Toss it 3 times. Record the outcomes.

HHH = Number 1

HHT = Number 2

HTH = Number 3

HTT = Number 4

THH = Number 5

THT = Number 6

TTH = Reject it, don’t ccount the toss and toss again

TTT = Reject it, don’t ccount the toss and toss again

Voila done!

What is the probability of HHH in this experiment?

Let X be the outcome in the restricted experiment as shown.

How is this experiment is different from the actual experiment?

This experiment is conditioning on the event A = {HHH, HHT, HTH, HTT, THH, THT}.

\(P( X = HHH) = P (X = HHH | X \in A ) = \frac{P (X = HHH)}{P (X \in A)} = \frac{1}{6}\)


Beautiful right?

Can you generalize this idea?

Food for thought

  • Give an algorithm to simulate any conditional probability.
  • Give an algorithm to simulate any event with probability \(\frac{m}{2^k}\), where \( m \leq 2^k \).
  • Give an algorithm to simulate any event with probability \(\frac{m}{2^k}\), where \( n \leq 2^k \).
  • Give an algorithm to simulate any event with probability \(\frac{m}{n}\), where \( m \leq n \leq 2^k \) using conditional probability.

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Logarithms and Equations | AIME I, 2000 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 2000 based on Logarithms and Equations.

Logarithms and Equations – AIME I 2000


\(log_{10}(2000xy)-log_{10}xlog_{10}y=4\) and \(log_{10}(2yz)-(log_{10}y)(log_{10}z)=1\) and \(log_{10}(zx)-(log_{10}z)(log_{10}x)=0\) has two solutions \((x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2})\) find \(y_{1}+y_{2}\).

  • is 905
  • is 25
  • is 840
  • cannot be determined from the given information

Key Concepts


Logarithms

Theory of Equations

Number Theory

Check the Answer


Answer: is 25.

AIME I, 2000, Question 9

Polynomials by Barbeau

Try with Hints


First hint

Rearranging equations we get \(-logxlogy+logx+logy-1=3-log2000\) and \(-logylogz+logy+logz-1=-log2\) and \(-logxlogz+logx+logz-1=-1\)

Second Hint

taking p, q, r as logx, logy and logz, \((p-1)(q-1)=log2\) and \((q-1)(r-1)=log2\) and \( (p-1)(r-1)=1\) which is first system of equations and multiplying the first three equations of the first system gives \((p-1)^{2}(q-1)^{2}(r-1)^{2}=(log 2)^{2}\) gives \((p-1)(q-1)(r-1)=+-(log2)\) which is second equation

Final Step

from both equations (q-1)=+-(log2) gives (logy)+-(log2)=1 gives \(y_{1}=20\),\(y_{2}=5\) then \(y_{1}+y_{2}=25\).

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Algebra Arithmetic I.S.I. and C.M.I. Entrance ISI Entrance Videos

Sum of divisors and Integers | TOMATO B.Stat Objective 99

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Sum of divisors and Integers.

Sum of divisors and Integers (B.Stat Objective Question)

The sum of all positive divisors of 1800, where 1 and 1800 are also considered as divisors of 1800, is

  • 104
  • 6045
  • 1154
  • none of these

Key Concepts


Integers

Sum of divisors

Exponents

Check the Answer


Answer: 6045

B.Stat Objective Problem 99

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

here 1800=(2)(2)(2)(3)(3)(5)(5) where n=\((p_1^a)(p_2^b)(p_3^c)\)

Second Hint

sum of divisors of 1800

=\((\frac{2^{4}-1}{2-1})\)\((\frac{3^{3}-1}{3-1})\)\((\frac{5^{3}-1}{5-1})\) where sum of divisors of n=\((\frac{p_1^{a+1}-1}{p_1-1})(\frac{p_2^{b+1}-1}{p_2-1})(\frac{p_3^{c+1}-1}{p_3-1})\)

Final Step

=6045.

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Algebra Arithmetic I.S.I. and C.M.I. Entrance ISI Entrance Videos

Divisibility and Integers | TOMATO B.Stat Objective 89

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Integers and divisibility.

Divisibility and Integers (B.Stat Objective Question )


300 digit number with all digits equal to 1 is

  • divisible neither by 37 nor by 101
  • divisible by both 37 and 101
  • divisible by 37 and not by 101
  • divisible by 101 and not by37

Key Concepts


Integers

Remainders

Divisibility

Check the Answer


Answer: divisible by 37 and 101

B.Stat Objective Problem 89

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

here we take 300 digit number all digit 1s

Second Hint

111…11=\(\frac{999…99}{9}\)(300 digits)

=\(\frac{10^{300}-1}{9}\)=\(\frac{(10^{3})^{100}-1}{9}\)=\(\frac{(10^{3}-1)X}{9}\)

since \(10^{3}-1\)=999 is divisible by 37 then 111…11(300 digits) is divisible by 37

Final Step

111…11=\(\frac{999…99}{9}\)(300 digits)

=\(\frac{10^{300}-1}{9}\)=\(\frac{(10^{4})^{75}-1}{9}\)=\(\frac{(10^{4}-1)Y}{9}\)

since \(10^{4}-1\)=9999 is divisible by 101 then 111…11(300 digits) is divisible by 101.

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Algebra Arithmetic Combinatorics I.S.I. and C.M.I. Entrance ISI Entrance Videos

Sets and Probability | B.Stat Objective Problems

Try this problem from I.S.I. B.Stat Entrance Objective Problem from TOMATO based on Sets and Probability. You may use sequential hints to solve the problem.

Sets and Probability (B.Stat Objective problems)


Sixty students appeared in a test consisting of three papers I ,II, and III. Of these students, 25 passed in paper I, 20 in paper II and 8 in paper III. Further 42 students passed in at least one of papers I and II, 30 in at least one of papers I and III, 25 in at least one of papers II and III. Only one student passed in all the three papers. Then the number of students who failed in all the three papers is

  • 17
  • 15
  • 45
  • 33

Key Concepts


Sets

Probability

Algebra

Check the Answer


Answer: 15

B.Stat Objective Problem

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

Let failed P I=P(A)=35, p II=P(B)=40, p III =P(C)=52, P (IandII)=\(P(A \bigcap B)\)= 18 P (I and III)=\(P(A \bigcap C)\)=30 P (II and III)=\(P(B \bigcap C)\)=35 P (I or II or III)=\(P(A \bigcup B \bigcup C)\)=59

Second Hint

Then by Poincare Theorem, \(P(A \bigcup B \bigcup C)\)=P(A)+P(B)+P(C)-\(P(A \bigcap B)\)-\(P(A \bigcap C)\)-\(P(B \bigcap C)\)+x

Then 59=35+40+52-18-30-35+x where x is the required value

Final Step

then x=59-44=15.

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Algebra Arithmetic I.S.I. and C.M.I. Entrance ISI Entrance Videos

Number Series | B.Stat Objective Problem

Problem – Number Series ( B.Stat Objective Problem )


We are going to discuss about Number Series from B.Stat Objective Problem .

A student studying the weather for d days observed that(i) it rained on 7 days morning or afternoon, (ii) when it rained in the afternoon it was clear in the morning, (iii) there were five clear afternoons (iv) there were six clear mornings. Then d equals

  • 8
  • 9
  • 11
  • 10

Key Concepts


Series

Algebra

Integers

Check the Answer


Answer: 9

B.Stat Objective Problem

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

Here A1, A2, A3, A4, A5, A6, A7, A8, A9 are mornings of day 1, day 2, day 3, day 4, day 5, day 6, day 7, day 8 and day 9 again B1, B2, B3, B4, B5, B6, B7, B8, B9 are afternoons of day 1, day 2, day 3, day 4, day 5, day 6 and day 7, day 8 and day 9

A1 A2 clear morning B1 B2 rainy afternoon from first two conditions

Second Hint

A3 rainy A4 clear A5 rainy A6 clear A7 rainy and B3 clear B4 rainy B5 clear B6 rainy B7 clear from first two conditions

Final Step

So, from first 7 days, 4 clear mornings and 3 clear afternoons. Then A8 clear A9 clear B8 clear B9 clear from last two conditions .Then all condition satisfied that is 4+2=6 mornings clear 3+2 =5 afternoons clear. d=9 that is in 9 days.

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Algebra Arithmetic I.S.I. and C.M.I. Entrance ISI Entrance Videos

Numbers and Group | B.Stat Objective Problem

Problem – Numbers and Group (B.Stat Objective Problem)


We are going to discuss about Numbers and Group from B.Stat Objective Problem .

In a group of 120 persons there are 80 Bengalis, 40 Gujratis, Further 71 persons in the group are Muslims and the remaining Hindus, Then the number of Bengali Muslims in the group is

  • 8
  • more than 30
  • 11
  • 10

Key Concepts


number Series

Algebra

Integers

Check the Answer


Answer: more than 30

B.Stat Objective Problem

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

Assuming that out of 80 Bengalis, when 49 are Hindus then 31 are Muslims and when 9 are Hindus 71 are Muslims

Second Hint

Out of 71 Muslims when 40 are Gujratis 31 are Bengalis else 71 are Bengalis which contradicts with other options

Final Step

Then number of Bengali Muslims are from 31 to 71 that is more than 30.

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Algebra Arithmetic I.S.I. and C.M.I. Entrance ISI Entrance Videos

Greatest Integer and remainder | TOMATO B.Stat Objective 113

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Greatest Integer and remainder.

Greatest integer and remainders (B.Stat Objective Question)


The greatest integer which, when dividing the integers 13511, 13903 and 14593 leaves the same remainder is

  • 98
  • 2
  • 7
  • 56

Key Concepts


GCD

Greatest Integer

Remainder

Check the Answer


Answer: 2.

B.Stat Objective Problem 113

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

here all numbers are odd

Second Hint

13511, 13903 and 14593 leaves different remainders when divided by 98, 56 and 7

Final Step

13511, 13903 and 14593 leaves same remainder when divided by 2

then greatest integer 2.

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Number of divisors and Integer | B.Stat Objective | TOMATO 83

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Number of divisors and Integer.

Number of divisors and Integer (B.Stat Objective)


The smallest positive integer n with 24 divisors (where 1 and n are also considered divisors of n) is

  • 240
  • 360
  • 420
  • 480

Key Concepts


Number of divisors

Integer

Least positive integer

Check the Answer


Answer: 360

B.Stat Objective Question 83

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

number of divisors of 420=(2+1)(1+1)(1+1)(1+1)=24

Second Hint

number of divisors of 240=(4+1)(1+1)(1+1)=20

Final Step

number of divisors of 360=(3+1)(2+1)(1+1)=24 and number of divisors of 480=(5+1)(1+1)(1+1)=24 then required number =360.

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Algebra Arithmetic Combinatorics I.S.I. and C.M.I. Entrance ISI Entrance Videos

Combinatorics and Integers | TOMATO B.Stat Objective 93

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Combinatorics and Integers.

Combinatorics and Integers (B.Stat Objective Question)


The highest power of 18 contained in \({50 \choose 25}\) is

  • 104
  • 1
  • 1154
  • none of these

Key Concepts


Integers

Combinatorics

Exponents

Check the Answer


Answer: 1

B.Stat Objective Problem 93

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

here \({50 \choose 25}\)=\(\frac{50!}{(25!)^{2}}\)=\(\frac{(50)(49)(….)(26)}{(25)(24)(…)(1)}\)

Second Hint

\(=(2)^{13}(49)(47)(45)(43)(41)(39)(37)(35)(33)(31)(29)(27) \times \frac{1}{12!}\)

Final Step

\(=(2)^{10}(49)(47)(15)(43)(41)(13)(37)(35)(11)(31)(29)\times \frac{1}{(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}[(2)^{3}(27)(9)(3)]\)

\(=(2)^{10}(49)(47)(15)(43)(41)(13)(37)(35)(11)(31)(29)\times \frac{1}{(12)(11)(10)(8)(7)(5)(4)(1)}[(2)(9)]\)gives a factor of \((18)^{1}\) then highest power of 18 is 1.

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