PRACTICE QUESTIONS ON PROBABILITY

Question 1 :

Can two events be mutually exclusive and independent simultaneously?

Solution :

If two events A and B are mutually exclusive then 

P(AnB)  =  0    ----(1)

If two events A and B are independent then 

P(AnB)  =  P(A) ⋅ P(B)    ----(2)

(1)  =  (2)

P(A) ⋅ P(B)  =  0

Case 1 :

If either P(A) or P(B) be equal to 0, then the events will be mutually exclusive or independent.

Case 2 : 

If P(A) > 0 and P(B) > 0, then 

P(AnB)  =  P(A) P(B)  > 0 (Which is not equal to 0)

So, the events are independent not mutually exclusive.

Case 3 : 

If P(A) > 0 and P(B) = 0 or P(A) = 0 and P(B) > 0, then 

P(AnB)  =  P(A) P(B)  = 0 

So, the events are mutually not independent exclusive.

Hence, we may conclude that two events cannot be mutually exclusive and independent simultaneously.

Question 2 :

If A and B are two events such that P(A U B) = 0.7, P(A n B) = 0.2, and P(B) = 0.5, then show that A and B are independent.

Solution :

To show that event A and B are independent, we have to show that P(AnB)  =  P(A)  P(B)

P(AUB)  =  P(A) + P(B) - P(AnB)

0.7  =  P(A) + 0.5 - 0.2

0.7  =  P(A) + 0.3

0.7 - 0.3  =  P(A)

P(A)  =  0.4

P(AnB)  =  P(A)  P(B)

0.2  =  0.4 (0.5)

0.2  =  0.2

Hence the events A and B are independent.

Question 3 :

If A and B are two independent events such that P(A∪B) = 0.6, P(A) = 0.2, find P(B).

Solution :

Since A and B are independent events,

P(AUB) = P(A) x P(B)

P(A∪B) = 0.6, P(A) = 0.2

Since A and B are independent events, P(AnB)  =  0

P(AUB)  =  P(A) + P(B) - P(AnB)

P(AUB)  =  P(A) + P(B) - [P(A) x P(B)]

0.6 = 0.2 + P(B) ( 1 - P(A) )

0.4 = P(B) (1 - 0.2)

P(B) = 0.4/0.8

P(B) = 4/8

P(B) = 0.5

Question 4 :

If P(A) = 0.5, P(B) = 0.8 and P(B/A) = 0.8, find P(A / B) and P(A∪B) .

Solution :

P(A/B)  =  P(AnB)/P(B)   ----(1)

P(B/A)  =  P(AnB)/P(A)

  0.8  =  P(AnB)/0.5

  0.8(0.5)  =  P(AnB)

  P(AnB)  =  0.4

By applying the value of P(AnB) in (1), we get

P(A/B)  =  0.4 / (0.8)

P(A/B)   =  0.5

P(AUB)  =  P(A) + P(B) - P(AnB)

P(AUB)  =  0.5 + 0.8 - 0.4

P(AUB)  =  0.9

Question 5 :

If for two events A and B, P(A) = 3/4, P(B) = 2/5 and AUB = S (sample space), find the conditional probability P(A/B)

Solution :

P(A) = 3/4, P(B) = 2/5 and AUB = S

P(AUB)  =  P(A) + P(B) - P(AnB)

1  =   (3/4) + (2/5) - P(AnB)

1  =  (15 + 8)/20 - P(AnB)

  1  -  (23/20)  =  P(AnB)

  P(AnB)  =  3/20

P(A/B)  =  P(AnB)/P(B)

  =  (3/20) / (2/5)

  =  (3/20) ⋅ (5/2)

P(A/B)  =  3/8

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Dec 23, 24 03:47 AM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 91)

    Dec 23, 24 03:40 AM

    Digital SAT Math Problems and Solutions (Part - 91)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 90)

    Dec 21, 24 02:19 AM

    Digital SAT Math Problems and Solutions (Part - 90)

    Read More