Blog: Mth202 Assignment No. 5 Solution

Assignment 5 Of MTH202 (Fall 2010)

Maximum Marks: 15                                                                                       

Due Date: January 24, 2011

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Question 1;                                                                                                            Mark: 05

Let A and B are subsets of U with n(U) = 200, n(A) = 30, n(B) = 90 and

n() = 120. Find n().

Question 2;                                                                                                          Marks: 04

In how many ways can six different books be arranged on a shelf?

Question 3;                                                                                                          Marks: 06

A pair of fair dice is rolled once. Find the probability that the sum of dots on the faces is 9.

....................

SOLUTION:

Question # 01

Let A and B are subsets of U with n (U) = 200, n (A) = 30, n (B) = 90 and

A

/

)

( UB

AB

∩

n(

Solution:

(A U B)´ = U \ (A U B)

∪

∪

n ((A B) ´) = n (U) – n (A

B)

∪

120 = 200 – n (A

B)

∪

n (A

B) = 200 – 120 = 80

Now by applying inclusion exclusion principle

∪

∩

n (A

B) = n (A) + n (B) – n (A B)

∩

80 = 30 + 90 - n (A B)

∩

n (A B) = 30 + 90 – 80 = 40

Question # 02

In how many ways can six different books be arranged on a shelf?

Solution:

) = 120. Find n (

).

We seek the no. of ways in which six different can be arranged on a shelf.

So

=

!=

xxxxx

11

6654321

720

Therefore six different books can arranged on a shelf in 720 different ways.

Question # 03

A pair of fair dice is rolled once. Find the probability that the sum of dots on the faces is 9.

Solution:

When a pair of dice is rolled its sample space S has 36 outcomes

S = {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}

Suppose E is the event that the sum of dots on faces is 9.

E = { (3, 6) (4, 5) (5, 4) (6, 3)}

So

===

nS

pE

nE

()369

()

()41

Complete