Mathematics

Conditional probability of events A and B $P(A\mid B) = \frac{P(A\cap B)}{P(B)}$

Bayes Theorem

 $\begin{aligned} P(A \mid B) &= \frac{P(A\cap B)}{P(B)}, \text{ if } P(B) \neq 0, \\ P(B \mid A) &= \frac{P(B\cap A)}{P(A)}, \text{ if } P(A) \neq 0, \\ \Rightarrow P(A\cap B) &= P(A\mid B)\times P(B)=P(B\mid A)\times P(A), \\ \Rightarrow P(A \mid B) &= \frac{P(B \mid A) \times P(A)} {P(B)}, \text{ if } P(B) \neq 0. \end{aligned}$

A couple has two children, the older of which is a boy. What is the probability that they have two boys?

 $\begin{aligned} P(B) &= Older one is boy \\ P(A) &= Two boys \\ \\ \\ P(A|B) &= \frac{P(B/A) P(A)}{P(B)} \\ &=\frac{P(older~one~is~boy~given~two~boys) * P(having~two~boys)}{P(older~one~is~boy)} \\ Given Fact, \\ P(older~one~is~boy~given~two~boys) &= 1 \\ P(A|B) &= \frac{1 * 1/2*1/2}{1/2} \\ &= 1/2 \end{aligned}$

A couple has two children, one of which is a boy. What is the probability that they have two boys?

 $\begin{aligned} A &= having two boys\\ B &= one child is a boy\\ \\ \\ P(one~child~boy) &= 1-P(both~children~not~boys)\\ &= 1- 1/4 \\ &= 3/4\\ P(having~two~boys~given~one~child~boy) &= \frac{P(one~child~is~boy~gicen~both~children~boys) * P(both~children~boys)}{P(one~child~boy)} \\ &=\frac{1 * 1/4}{3/4} \\ &= 1/3\\ \end{aligned}$

A family has two children. Given that one of the children is a boy, and that he was born on a Tuesday, what is the probability that both children are boys?

 $\begin{aligned} P(BB) &= Probablity~of~having~both~boys~of~two~children \\ P(B) &= Probablity~of~having~one~boy~of~two~children\\ P(BT) &= probablity~of~having~a~boy~born~on~Tuesday \\ \\ P(BB|BT) &= \frac{P(BT|BB) *P(BB)}{ P(BT)} \\ P(BT|BB) &=P(having~boy~born~on~Tuesday~given~both~children~are~boys) \\ P(BT|BB) &=P(Either~of~two~can~be~boy~~born~on~Tuesday~given~both~boys) \\ \\ P(BT|BB) &= 1- P(boys~born~on~other~days)\\ &=1-6/7*6/7 \\ &= 1-36/49 = 13/49\\ \\ P(BB) &= 1/2*1/2 = 1/4 \\ \\ P(BT) &= P(one~of~the~child~is~a~boy~born~on~Tuesday) \\ &=1-P((both~boys~born~on~non-Tuesdays) + both~girl) \\ &= 1- 13/14 * 13/14 \\ &= 1- 169 /196 = 27/196\\ \\ P(BB|BT) &= P(BT|BB) *P(BB)/ P(BT) \\ &= 13/49 * 1/4 / 27/196 \\ &=13/49 * 1/4 * 196 /27 \\ &= 13/27 \end{aligned}$

Balls numbered 1 through 20 are placed in a bag. Three balls are drawn out of the bag without replacement. What is the probability that all the balls have odd numbers on them

Three balls drawn with out replacement

Total Probability = P(1st ball) P(2nd ball) P(3rd ball)

= 10/20 9/19 8/18

In how many possible ways can you write 1800 as a product of 3 positive integers a, b, c ?

Prime Factorization of 1800 - $2^3 * 3^2 * 5^2$
Distribute power of prime factors among digits a, b,c as asked. For example

If we distribute 2^4 as (2,2,0) and, lets say a,b,c with powers of other prime factors 3 and 5 are (1, 9,25) ,then, with powers of 2 distributed as (2,2,0) (total 4), becomes

$(2^2 *1, 2^2 * 9, 2^0 * 25) = (4, 36,25 )= 4*36*25 = 1800$

Take another example

Distribute as (0,3,1)

$(2^0 * 1, 2^3 * 9, 2^1 * 25) = (1, 72,25) = 1800$

  2^4 can be distibuted like the followings among 3 numbers (a,b,c). Smilarly powers of other prime factors. Together the product is 1800
  4,0,0
  0,4,0
  0,0,4
  3,1,0
  3,0,1
  1,3,0
  1,0,3
  ......

  2,2,0
  2,0,2
  0,0,2
  2,1,1
  1,2,1
  1,1,2
  ...etc. etc

So problem becomes distributing power r (4 here for prime factor 2) among n (3 here) numbers.

Distribute r items among n people

 $D_(r,n)=\frac{(r+ (n-1))!}{r! (n-1)!}$

Find this sequence of combinations $D_(r,n)$ in Pascals triangle

Answer is $D_{(3,3)} * D_{(2,3)} * D_{(2,3)} = 10 * 6 * 6$

Ans: 360

How many ways letters in word ARMOUR can be arranged

n = 6

repeating letters = R -2 times, Combinations of them are indistinguishable so will have to reduce from total permutations

So net permutations = $\frac{n!}{r!}$ = 720 / 2 = 360

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ?

Number of ways 3 consonants and 2 vowels \ can be selected =

here combination is taken because order of letters not important. However order is important when you form words with these selected letters

 $=\frac{7!}{(7-3)! 3!} * \frac{4!}{(4-2)! 2!} \\ =35 * 6\\ = 210 \\$

Permutations of words with 3 consonants and 2 vowels

= $5!$ = 120

So total combinations = $120 * 210 = \underline{25200}$

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated ?

Divisible by 5 means last digit is 5.

For remaining 2 digits, total permutations without repeat =

$= {^5\!P_2}$

$=\underline20$

Evaluate permutation equation 59P3 ?

= 59 X 58 X 57

How many words can be formed from the letters of the word 'AFTER', so that the vowels never come together ?

For 5 letters, total permutations 5!

Consider A and E as one letter , ie. total 4 letters and then permutations would be 4!

AE can have two arrangements, AE and EA= 2!

So total permutations with AE and EA considered one letter = 2 * 4!

So total permutations where A and E doesn't come together

= 5! - 2*4! = 72

A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colors ?

Since there are three colors and three draws, different color should come in each draw, so total number ways three different colors can be draws in three draws = 4 3 2 = 24

In how many different ways the letters of the word "MATHEMATICS" can be arranged so that the vowels always come together ?

Total 11 letters

Vowels - AEAI. A repeats

Consonants M and T repeats

Consider vowels AEAI as one letters =V, since it is asked to find arrangements where they come together. However A repeats so, combination needs to divide by 2!

Total number of letters with AEAI together as one letter = 8. Total arrangements = 8!. However M and T repeats . So arrangements become 8!/(2!*2!)

Now with AEAI, they can have their own arrangements, still being together. So those arrangements are 4!/2! (since A repeats)

total permutations= 8! x 4!/8=7! * 4! = 120960

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw ?

Total Combinations of 9 balls into 3= ${}^9C_3$

Total combination where black ball doesn't show up = ${}^{(number of other balls)}C_3 = []^6C_3$

Number of combinations where at least one black ball shows up = ${}^9C_3 - {}^6C_3$

= 84-20 = 64

How many groups of 6 persons can be formed from 8 men and 7 women ?

${}^{15}C_3$

8 men entered a lounge simultaneously. If each person shakes hand with the other, then find the total no. of handshakes?

shake hand is group of 2

No order, A shakes B same as B shakes A

So ${}^8C_2$

In how many different ways, can the letters of the word 'INHALE' be arranged ?

INHALE has 6 letters, none repeating.

Total arrangements = 6!

In how many different ways, can the letters of the word 'BANKING' be arranged ?

BANKING has 7 letters, N repeating

Total Arrangements =

 $\frac{7!}{2!}$

In a meeting between two countries, each country has 12 delegates. All the delegates of one country shake hands with all delegates of the other country. Find the number of handshakes possible ?

all x all = 12 x 12

How many can straight lines be drawn from 15 non-collinear points ?

Combination of 2 points. And a line cant be drawn from a point to itself

$={}^{15}C_2 - 15$

=105

There are 25 students in a class. Find the number of ways in which a committee of 3 students is to be formed?

Combination as order is not important

${}^{25}C_3$

In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all person of the same nationality sit together?

8I, 4A, 4E

Consider each group as whole and find their seating

3 groups, order important, 3P3 = 3!

In each group, you can have arrangements, order

Total =3! 8! 4! * 4!

How many different words can be formed using all the letters of the word ALLAHABAD?
(a) When vowels occupy the even positions.
(b) Both L do not occur together.

9 letters,

Repeats A -4, L - 2,

Vowels = A only, So only one vowel combination

Remaining 5 letters, and arrangements = 5!

Consider both L's together as single letter, then arrangements = 4!

So arrangements where both Ls don't occur together = 5!-4!

120 - 24= 96

Revise Conditional Probability and Bayes Theorem

Revise Permutations and Combinations