PROBLEMS ON PROBABILITY WITH OR WITHOUT REPLACEMENT

Problem 1 :

A bag contains 12 white balls and 16 black balls. If two balls are drawn is succession without replacement then what is the probability that the first is white and the second is black? 

a) 32/20   b) 43/63    c)  41/20    d)  16/63     e) 15/79

Solution :

Total number of balls = 12 white + 16 black

= 28 balls

Let A and B be the events of getting choosing firts and second balls respectively.

P(A) = n(A)/n(S)

= 12/28

Since the replacement is not done, then the total number of balls will be reduced by 1.

P(B) = 16/27

= (12/28) x (16/27)

= 16/63

Problem 2 :

A bag contains 4 red balls, 6 blue balls and 11 black balls. If two balls are drawn one by one without replacement, find the probability that two balls are black.

a) 4/21    b) 3/7    c)  11/42    d) 8/69    e) none

Solution :

Total number of balls = 4 red + 6 blue + 11 black

= 21 balls

Let A be the event of getting black balls.

P(AA) = P(A) x P(A)

Replacement is not done, then the total number of balls is reduced by 1.

= 11/21 x 10/20

= 11/42

So, option c is correct.

Problem 3 :

A box contains 10 pencils m erasers and n pens and the probability of getting an eraser is 3/10 and the probability of getting pencil is 0.5. If three products are drawn one by one without replacement, then find the probability that the first one is pen, the second one is pencil and the third one is pen.

a)  1/57    b) 4/81    c) 13/64   d) 5/72     e) none

Solution :

Total number of items = 10 pencils + m erasers + n pens

= 10 + m + n

Probability of getting an eraser = 3/10

m/(10+m+n) = 3/10

10m = 3(10 + m + n)

10m = 30 + 3m + 3n

 10m - 3m - 3n = 30

7m - 3n = 30 ----(1)

Probability of getting pencil = 0.5

= 5/10

= (5 x 2)/(10 x 2)

= 10/20

Total number of pencils / total number of items

Total number of items = 10 + m + n

20 = 10 + m + n

m + n = 10 -----(2)

(1) + (2) 3 ==> 7m - 3n + 3m + 3n = 30 + 30

10m = 60

m = 6

Applying m = 6, we get

n = 10 - 6

n = 4

Probability of getting pens = 4/20

Probability of choosing three items (first pen, second pencil, third pen) 

= (4/20) x 10/19 x (3/18)

= 1/57

Problem 4 :

A box contains 7 yellow shirts 4 white shirts and some orange shirts and the probability of getting orange shirt is 5/16. If three shirt are drawn one by one without replacement, then find the probability of that three shirts are in different color.

Solution :

Let x be the orange number of shirts.

Total number of shirts = 7 yellow + 4 white + x orange

= 11 + x

probability of orange shirts = 5/16

x/(11 + x) = 5/16

16x = 5(11 + x)

16x = 55 + 5x

16x - 5x = 55

11x = 55

x = 55/11

x = 5

Probability of getting shirts of difference color = 7/16 x (4/15) x (5/14)

= 1/24

Problem 5 :

There are 20 apples and 10 mangoes in the box. If two fruits are drawn by one without replacement in the box, then the probability that at most is an apple.

a)  5/8       b) 1/3    c)  1/7    d)  2/5

Solution :

Total number of fruits = 20 apples + 10 mangoes

= 30

Choosing one apple + choosing no apples

= 20/30 x 10/29 + 10/30 x 9/29

= 20/87 + 9/87

= (20 + 9)/87

= 29/87

= 1/3

So, option b is correct.

Problem 6 :

A box contains orange, yellow  and blue balls in the ratio 6 : 2 : 5 respectively and sum of the total number of orange and blue balls in the box is 36 more than that of yellow balls. If three balls are drawn one by one without replacement, then find the probability that three balls are blue.

a)  21/287       b) 92/2002    c)  57/1105    d)  32/1989

Solution :

Total number of balls = 6x orange + 2x yellow + 5x blue

number of orange and blue balls = 36 + number of yellow balls

6x + 5x = 36 + 2x

11x - 2x = 36

9x = 36

x = 36/9

x = 4

Number of orange balls = 6(4) ==> 24

Number of yellow balls = 2(4) ==> 8

Number of blue balls = 5(4) ==> 20

Total number of balls = 24 + 8 + 20

= 52

Probability of getting three blue balls = 20/52 x 19/51 x 18/50

= 2/13 x 2/13 x 2/13

= 57/1105

So, option c is correct.