Solutions for Probability

1. Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) =	n(E) /	=	9 /	.
	n(S)		20

2. Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S)	= Number of ways of drawing 2 balls out of 7
	= 7C2 `
	= (7 x 6) / (2 x 1)
	= 21.

Let E = Event of drawing 2 balls, none of which is blue.

n(E)	= Number of ways of drawing 2 balls out of (2 + 3) balls.
	= 5C2
	= (5 x 4) / (2 x 1)
	= 10.

P(E) =	n(E) /	=	10 /	.
	n(S)		21

3. Total number of balls = (8 + 7 + 6) = 21.

Let E	= event that the ball drawn is neither red nor green
	= event that the ball drawn is blue.

n(E) = 7.

P(E) =	n(E) /	=	7 /	=	1 /	.
	n(S)		21		3

4. In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) =	n(E) /	=	4 /	=	1 /	.
	n(S)		36		9

5. Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) =	n(E) /	=	7 /	.
	n(S)		8

6.  Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S)	= Number ways of selecting 3 students out of 25
	= 25C3 `
	= (25 x 24 x 23) (3 x 2 x 1)
	= 2300.

n(E)	= (10C1 x 15C2)
	= 10 x (15 x 14) / (2 x 1)
	= 1050.

P(E) =	n(E) /	=	1050 /	=	21 /	.
	n(S)		2300		46

7.  Let S be the sample space.

Then, n(S) = 52C2 =	(52 x 51) /	= 1326.
	(2 x 1)

Let E = event of getting 2 kings out of 4.

n(E) = 4C2 =	(4 x 3) /	= 6.
	(2 x 1)

P(E) =	n(E) /	=	6 /	=	1 /	.
	n(S)		1326		221

8.  Let S be the sample space.

Then, n(S)	= number of ways of drawing 3 balls out of 15
	= 15C3
	= (15 x 14 x 13) / (3 x 2 x 1)
	= 455.

Let E = event of getting all the 3 red balls.

n(E) = 5C3 = 5C2 =	(5 x 4) /	= 10.
	(2 x 1)

P(E) =	n(E) /	=	10 /	=	2 /	.
	n(S)		455		91

9.  Let S be the sample space.

Then, n(S) = 52C2 =	(52 x 51) /	= 1326.
	(2 x 1)

Let E = event of getting 1 spade and 1 heart.

n(E)	= number of ways of choosing 1 spade out of 13 and 1 heart out of 13
	= (13C1 x 13C1)
	= (13 x 13)
	= 169.

P(E) =	n(E) /	=	169 /	=	13 /	.
	n(S)		1326		102

10.  A number divisible by 4 formed using the digits 1,2,3,4 and 5 has to have the last two digits       12 or24 or 32 or 52.

In each of these cases, the five digits number can be formed using the remaining 3 digits in 3! = 6 ways.

A number divisible by 4 can be formed in 6×4=24 ways.

Total number that can be formed using the digits 1,2,3,4 and 5 without repetition
=5!=120

Required probability = 24/120 = 1/5

11.  Let A be the event that ball selected from the first bag is red and ball selected from second bag is green.

Let B be the event that ball selected from the first bag is green and ball selected from second bag is red.

P(A)=(5/8) * (6/10)         = 3/8 And P(B) =( 3/8 ) * (4/10)          = 3/ 20 Hence required probability = P(A) + P(B)

= 3/8 + 3 /20

= 21/40

12. Total cases of checking in the hotels =43 ways.

Cases when 3 men are checking in different hotels =4×3×2=24 ways.

Required probability:  = 24/43
                                        = 3/8

13.    n(S) =4×5=20
          n(E)   = 4
       P(E)  = n(E) / n(S)

                = 4 / 20

                 = 1 / 5

                 = 0.20

14.  The events of selection of two person is redefined as (i) first is a girl AND second is a boy OR (ii) first is boy AND second is a girl OR (iii) first is a girl and second is a girl.

So the required probability:=( 9/17 * 8/ 16 ) + ( 8 /17 * 9/16 ) + (9/17 * 8/16 ) =9/ 34 + 9 / 34+ 9 / 34                                                     =  25 / 34

15.  n(S)= number of ways of sitting 12 persons at round table:
=(12−1)!=11!

Since two persons will be always together, then number of persons:
=10+1=11

So, 11 persons will be seated in (11−1)!=10! ways at round table and 2 particular persons will be seated in 2! ways.

n(A)= The number of ways in which two persons always sit together =10!×2

P(A)=n(A) /n(S)          = 10! * 2! /11!          = 2 / 11

16.  The total number of words which can be formed by permuting the letters of the word 'UNIVERSITY' is 

10! / 2!

as there is two I's.

Hence n(S)= 10! / 2!

Taking two I's as one letter, number of ways of arrangement in which both I's are together = 9!
So n(X)=9!

Hence required probability =n(X)/ n(S)                                                 = 9! / ( 10! 2!)                                                 = 1/5

17.  Total No.of Balls = 12

            n(S) = 12C₂ = 66

            n(E)= 4C₂ + 3C₂+ 5C₂ = 19

            P(E)= n(E)/n(S) = 19/66

18.  Selection of 1 boy and 3 girls in 5C₁ * 4C₃ = 20 ways

     Selection of 4 girls and no boy in 5C₀ * 4C₄ = 1 way

     Therefore n(E) = 21 ways

  4 girls and 5 boys in 9C₄ = 126 ways

     p(E) = n(E) / n(S) = 21/126 = 1/6

19. There are eight different letters in the given word .

              Total no.of arrangements n(S) = 8 P ₈ = 8 !

              If N occupies first place , E must occupy  last place and vice versa so that there are exactly six letters in between the letters N and E

            N and E can be arrange in 2P₂ = 2!

And rest six places can be filled by remaining six letters ( Q U A T I and O ) in 6P₆ = 6!

                        n(E)= 2! * 6!

                        p(E) = n(E) / n(S) = (2 * 6)! /  8!

                                    = 1/ 28

20. A three digit number can be formed with a given 5 digit in 5P₃ ways

                           n(S) = 5P₃ = 60

            any one of two digits 4 and 6 should come at unit place which can be done in 2 ways. 

After filling up the unit place the remaining two places can be filled up with the remaining four digits in 4P₂ ways

                                    n(E) = 2 * 4P₂ = 24

                        p(E) = n(E) / n(S) = 24/60= 2/5