PRACTICE PROBLEMS ON PROBABILITY WITH SOLUTION

Problem 1 :

Two balls are drawn from a bag containing 5 white and 7 black balls at random. What is the probability that they would be of different colors ?

Solution :

Total number of balls  =  5 white + 7 black

n(s)  =  12 balls

Let A be the event of selecting white ball and B be the event of selecting black ball.

n(A)  =  5 and n(B)  =  7

p(A)  =  n(A)/n(S)  =  5/12

p(B)  =  n(B)/n(S)  =  7/12

P(AB) + p(BA)  =  (5/12) x (7/11) +  (7/12) x (5/11)

=  35/132 + 35/132

=  70/132

=  35/66

Problem 2 :

What is the chance of throwing at least 7 in a single cast with 2 dice ?

Solution :

Sample space  =  {(1, 1) (1, 2) (1, 3) ..............(6, 6)}

n(S)  =  36

Sum 7 :

A  =  { (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) }

n(A)  =  6

Sum 8 :

B  =  { (2, 6) (3, 5) (5, 3) (4, 4) (6, 2) }

n(B)  =  5

Sum 9 :

C  =  { (3, 6) (4, 5) (5, 4) (6, 3) }

n(B)  =  4

Sum 10 :

D  =  {(4, 6) (5, 5) (6, 4) }

n(D)  =  3

Sum 11 :

D  =  { (5, 6) (6, 5) }

n(D)  =  2

Sum 12 :

E  =  { (6, 6) }

n(E)  =  1

P(Getting at least 7 in a single cast)  =  (6+5+4+3+2+1)/36

=  21/36

=  7/12

Problem 3 :

What is the chance of getting at least one defective item if 3 items are drawn randomly from a lot containing 6 items of which 2 are defective item ?

Solution :

Total number of items  n(s)  =  6 C ₃

=  6⋅5⋅4/(3⋅2⋅1)

=  20

Number of defective items  =  2

P(choosing at least one defective item)  :

Number of choosing 1 defective item  =  4C₂ x 2C₁

Number of choosing 2 defective items  =  4C₁ x 2C₂

Number of choosing at least one defective item :

=  (4C₂ x 2C₁) + (4C₁ x 2C₂)

=  12 + 4

=  16

Required probability  =  16/20  ==>  0.80

Problem 4 :

If two unbiased dice are rolled together, what is the probability of getting no difference of points ?

Solution :

Sample space  =  n(S)  =  36

No difference  =  0 difference 

Let A be the event of getting 0 difference.

A  =  {(1, 1)(2, 2)(3, 3) (4, 4) (5, 5) (6, 6)}

n(A)  =  6

p(A)  =  n(A)/n(S)

  =  6/36

p(A)  =  1/6

Problem 5 :

If A, B and C are mutually exclusive independent and exhaustive events then what is the probability that they occur simultaneously ?

Solution :

What is mutually exclusive ?

A set of two or more events are said to be mutually exclusive if and only if it is impossible for the events to occur together. That is the definition of the term mutually exclusive. So obviously the probability of a set of mutually exclusive events occurring simultaneously is zero.

Problem 6 :

There are 10 balls numbered from 1 to 10 in a box. If one of them selected at random. What is the probability that the number printed on the ball would be an odd number greater than 4 ?

Solution :

Sample space n(S)  =  10

Let A be the event getting odd number greater than 4.

A  =  {5, 7, 9}

n(A)  =  3

p(A)  =  n(A)/n(S)

p(A) = 3/10

Problem 7 :

When 2 fair dice are thrown what is the probability of getting the sum which is multiple of 3 ?

a)  4/36    b)  13/36    c)  2/36    d)  12/36

Solution :

If two dice are rolled, then the sample space = 36

Let A be the event of getting the sum which is multiple of 3. {3 or 6 or 9 or 12}

A = {(1, 2) (2, 1) (2, 4) (4, 2) (3, 3) (5, 1) (1, 5) (6, 3) (3, 6) (4, 5) (5, 4) (6, 6)}

n(A) = 12

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

= 12/36

So, option d is correct.

Problem 8 :

Which two coins are tossed simultaneously the probability of getting at least one tail ?

a)  1   b)  0.75     c)  0.5    d)  0.25

Solution :

When two coins are tossed, the sample space

S = {HH, TT, HT, TH}

n(S) = 4

Let A be the event of getting at least one tail

A = {HT, TH, TT}

n(A) = 3

p(A) = n(A) / n(S)

= 3/4

= 0.75

Problem 9 :

A bag contains 3 white and 5 black balls and second bag contains 4 white and 2 black balls. If one ball is taken from each bag, the probability that both balls are white is ____

A)  1/3    b)  1/4    c)  1/2    d) None of these

Solution :

Number of balls in bag 1 :

3 white + 5 black = 8 balls

Number of balls in bag 2 :

4 white + 2 black = 6 balls

Probability of choosing 1 white ball from bag 1

= ³C₁ / ⁸C₁

= 3/8 -----(1)

Probability of choosing 1 white ball from bag 2

= ⁴C₁ / ⁶C₁

= 4/6 -----(2)

Since we choose 1 ball from each bag, it must be independent events.

Required probability = (3/8) x (4/6)

= 1/4

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