Probability

Pratyush @ Toughest Questions

Probability


What is probability?

favorable casestotal cases


In tossing a coin, events are:

P ( Head) = 1/2

P ( Tail ) = 1/2


Probability (Head or Tail)1/2 + 1/2 = 1

Probability ( A or B)

= P(A) + P(B)

Probability 1


Example of 2 coins tossed

Case 1: 2 H

Case 2: 2 T

Case 3: H & T


What is the probability of “a head and tail”?

Favorable outcomeTotal outcome

= 1/3


Case 1: HHCase 2 : TTCase 3 : HTCase 4 : TH

So P = 2/4 = 1/2

Wrong


With 3 coins, what is the probability of only 1 head?P (HTT) = (1/2)*(1/2)*(1/2) = 1/8

Joint P = P(A) * P(B) * P(C)P (THT) and P (TTH) are also

1/8 each.

So P = 1/8 + 1/8 + 1/8 = 3/8


2 dice are thrown

Find the probability that the sum of faces is 1

P = 0

Probability 0



Probability = 1/36



Probability = 1/36



Total Sum Set = { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

So Probability = 1/11


WrongTotal outcomes = 6*6 = 36

Favorable outcomes { (3,3), (2,4), (1,5), (4,2), (5,1) }

P = 5/36

Probability of all outcomes should be equal


2 dice are thrown

Find the probability that the sum is single digit


Prob ( Sum = 2) = 1/36

Prob ( Sum = 3) = 2/36

Prob ( Sum = 4) = 3/36

Prob ( Sum = 5) = 4/36


Prob ( Sum = 6) = 5/36

Prob ( Sum = 7) = 6/36

Prob ( Sum = 8) = 5/36

Prob ( Sum = 9) = 4/36

Total = (1 + 2 + 3 + 4 + 5 + 6 + 5 + 4)

36


Alternatively

Prob ( Sum = 10) = 3/36

Prob ( Sum = 11) = 2/36

Prob ( Sum = 12) = 1/36


P ( Sum 10,11 or 12) = 6/36

P (Sum is not 10, 11 or 12) = 1 - 6/36 = 30/36

P(Ac) = 1 - P(A)Its easier to work out the complement sometimes...


4 dice are thrown -

probability that there are 3 sixes and another number

Favorable outcome = 1*1*1*5*4Total outcomes = 6*6*6*6

P = 20 / 64


Probability that a number is divisible by 2 is

P = 1/2


Probability that a number is divisible by 3 is

P = 1/3


Probability that a number is divisible by 2 or 3 is

P = 1/2 + 1/3 = 5/6


Is it?Probability that number is

divisible by 2 or 3 is 1/2 + 1/3 - 1/6 = 4/6 = 2/3

P (A B) = P(A) + P(B) - P(A B)


10 coins are tossed, what is the probability that only

one is a tail or head?


Prob of 1 H, 9 T is = 1/ 210, but there are 10 such cases. So

prob = 10/210

Prob of 9 H, 1 T is = 10/210

Prob of one tail or head is = 10/210 + 10/210 = 10/29


In a single toss of 4 coins what is the probability of

getting at least one head?

P (no heads) = 1/16.

P (at least 1 head) = 1 - 1/16 = 15/16


In a single toss of 4 coins what is the probability of getting at most 1 head?

P (no heads) = 1/16.P (1 H & 3 T) = 1/16 * 4.

P (at most 1 H) = P (0 H) + P (1 H) = 5/16


2 cards are successively drawn without

replacement. Probability that both are greater than

2 & less than 9 is


There are 6 * 4 = 24 cards

P (1st card) = 24/52

P (2nd card) = 23/51

Joint P = 24/52 * 23/51


Probability of 1st ball blue & 2nd white without

replacement?


first ball blue = 4/10 = 2/5.

second ball white = 6/9 = 2/3

Joint P = 2/5 x 2/3 = 4/15


Probability of all 4 being blue - without replacement?


= (7/13)*(6/12)*(5/11)*(4/10)

= 7/143


Probability of all four not blue - without replacement?


= 1 - 7/143

= 136/143


Probability that all are of same color?


All white = 6/13 x 5/12 x 4/11 x 3/10 = 3/143

Probability all same color =P (all B) + P (all W) = 10/143


The probability that a card drawn from a pack of 52 cards is a diamond/king?There are 13 diamonds and 4

kings, but one of them is a king of diamond. Therefore,

P = (13 + 4 - 1) / 52 = 16/52


In a lottery of tickets numbered 1 to 50, 2

tickets are drawn. The probability that both are

prime numbers is:


(15 prime numbers from 1 - 50)P (1st being prime) = 15/50 P (2nd being prime) = 14/49

P (both prime) = 15*14 / 50*49


An AA gun can take max 4 shots. P of hitting the plane at the 1st, 2nd, 3rd

and 4th shot are 0.4, 0.3, 0.2, 0.1 resp. P that the

gun hits the plane is:


Probability of hitting the plane =

P (the plane is hit at least once) = 1 - (1-p1)(1-p2)(1-p3)(1-p4) =1 - (0.6)*(0.7)*(0.8)*(0.9) =

(1 - 0.3024) = 0.6976


A contains 4 red and 5 white balls and B contains 3 red and 7 white balls. 1 ball is drawn from A and 2

from B. Find the probability that out of 3, 2

are white and 1 is red.


P(2 W & 1 R) = P[(1 R from 1st and 2 W from

2nd or(1 W from 1st,1 W and 1 R from

2nd)] (4/9)*(7/10)*(6/9)+ (5/9)*(7/9)*(3/10)

= 7/15


A speaks the truth in 75% and B in 80% of the

cases. In how many cases will they contradict each other narrating the same

incident.


P(A and B contradict each other)

= P[(A tells truth and B tells a lie) or

P(A tells a lie and B tells truth)= 3/4 * 1/5 + 1/4 * 4/5

= 7/20


P of two events E and F are 0.25 and 0.50 resp. P

of their simultaneous occurrence is 0.14. P that neither E occurs nor F is:


P (E U F) = 0.25 + 0.50 - 0.14

= 0.61. Now, P (neither E nor F)

= 1 - P(E U F) = 1 - 0.61 = 0.39


The probability that a leap year selected at random will contain 53 Sundays,

is:


A leap year contains 366 days and therefore, 52 weeks and 2

days. Now, there can be 7 different combinations of the remaining two days, where 2

favour the required event. Hence, P = 2/7.


The probability of solving a problem by A, B, and C

are 1/2, 1/3 and 1/4 respectively. The

probability that the problem is solved is:


P = 1 - P(that the problem remains unsolved)

= 1 - (1-p1)(1- p2)(1- p3)= 1 - (1/2)*(2/3)*(3/4)

= 1 - 1/4 = 3/4


If P that A and B will die within a year are a and b resp, P that only one of them will be alive at the

end of the year is:


Required probability = P(A dies & B alive) or

P(A is alive and B dies.) =a (1 - b) + (1 - a) b

= a + b - 2ab.


In a row of 6 Mathematics books and 4 Physics

books, the probability that 3 particular Maths books

will be together is:


Taking 3 books as one book, we have 8 books which can be

arranged in 8! ways.These 3 books can be

arranged in 3! ways. Hence n(F) = 8!*3! Also n(T) = 10!.

Therefore P = (8!*3!) / 10! = 1/15


N cadets stand in a row. If all possible permutations are equally likely, P of 2

particular cadets standing side by side is:

P = [(N-1)! * 2!] / N! = 2/N


A can solve 90% of the problems given in a book

and B can solve 70%. What is P that at least 1 of them will solve a problem,

selected at random?


P(E) = 0.9, P(F) = 0.7 Also P(EF) = P(E)*P(F) =

0.63.Hence required probability is

0.9 + 0.7 - 0.63 = 0.97


Bag X contains 3 B and 4 W balls. Bag Y contains 4 B and 2 W balls. A bag is selected at random. From the selected bag, 1 ball is drawn. Find P that it is W.


P (drawing W from X) = 4/7. P (drawing W from Y) = 2/6. Choosing either bag gives a

probability of 1/2. Hence, total probability

= 1/2 (4/7 + 2/6) = 19/42


3 urns contain 2 W & 3 B balls, 3 W & 2 B and 4 W

& 1 B resp. A ball is drawn from an urn at random

and it is W. What it P that it was drawn from 1st urn?


P = (2/5)*(1/3)

------------------------------------------(2/5)*(1/3) + (3/5)*(1/3) + (4/5)*(1/3)

= 2/9


X contains 5 W and 7 B balls. Y contains 7 W and 8

B. 1 ball is drawn from X and put into Y without

noticing its colour. If now a ball is drawn from Y, what is

P that it is W?


P(transfer W ball from X & then select W ball from Y) = 5/12* 8/16

= 5/24. P( transfer B ball from X and then draw B ball from Y) =

(7/12)*(7/16) = 49/192. Total probability is = (5/24) + (49/192) = 89/192

Probability

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