Mth202 Assignment No. 5 Solution
Friday, January 21, 2011 Posted In Mth Edit ThisAssignment 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
