# Thread: Assignment 5 Of MTH202 (Spring 2010)

1. ## Assignment 5 Of MTH202 (Spring 2010)

Maximum Marks: 15
Due Date: July 20, 2010

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Question 1; Marks: 10

There are 8 men and 10 women members of a club. How many committees of seven can be formed, having;
1) at the most 5 men
2) atleast 3 women

Question 2; Marks: 5

A box contains six discs numbered 1 to 6.Find for each integer k (from 4 to 10), the probability that the numbers on two discs drawn without replacement have a sum equal to k.

Code:
`plz solve the assignments`  Reply With Quote

2. last date is 19 plz solve it  Reply With Quote

3. plz koi idea provide kar dy  Reply With Quote

4. app logon ka tu yah subject hai why dont u discuss and try to solve.  Reply With Quote

5. please make the discussion and solve it  Reply With Quote

6. yes thats the right way to solve it. share ur knowladge  Reply With Quote

7. A box contains six discs numbered 1 to 6.Find for each integer k (from 4 to 10), the probability that the numbers on two discs drawn without replacement have a sum equal to k.  Reply With Quote

8. So what is the conclusion ?  Reply With Quote

9. first question to kar liya wo to is tariky sy karian gy
men=8
women=10
(1) ways of at the most 5 men= c(8,5).c(10,2) + c(8,4) . c(10,3) + c(8,3) .(10,4)+C(8,2).c(10,5)
+ (8,1) .c(10,6) +c(8,0) .(10,7)
=56*45 + 70 * 120 + 56*210 + 28*252 + 8*210 + 1*120
=2520+8400+11760+7056+1680+120
=31536
(2) ways of at least 3 women=c(8,4) .c(10,3) + c(8,3) .c(10,4) + c(8,2) .c(10,5) + c(8,1).c(10,6)
+c(8,0) .c(10,7)
= 8400+11760+7056+1680+120
=29016
pr ab 2nd question me koi ehlp kar dy plz  Reply With Quote

10. ## Idea Solution

Question 1;

There are 8 men and 10 women members of a club. How many committees of seven can be formed, having;
1)at the most 5 men
2)atleast 3 women

solution:

men=8
women=10
(1) ways of at the most 5 men= c(8,5).c(10,2) + c(8,4) . c(10,3) + c(8,3) .(10,4)+C(8,2).c(10,5)
+ (8,1) .c(10,6) +c(8,0) .(10,7)
=56*45 + 70 * 120 + 56*210 + 28*252 + 8*210 + 1*120
=2520+8400+11760+7056+1680+120
=31536
(2) ways of at least 3 women=c(8,4) .c(10,3) + c(8,3) .c(10,4) + c(8,2) .c(10,5) + c(8,1).c(10,6)
+c(8,0) .c(10,7)
= 8400+11760+7056+1680+120
=29016

Question 2
A box contains six discs numbered 1 to 6.Find for each integer k (from 4 to 10), the probability that the numbers on two discs drawn without replacement have a sum equal to k.
solution:

All posibale sample w,o,r, of size 2 are=(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5 ),(2,6),(3,4),(3,5),(3,6),
(4,5),(4,6),(5,6)

p(4)=1/15
p(5)= 2/15
p(6)=2/15
p(7)=3/15
p(8)=2/15
p(9) = 2/15
p(10)=1/15  Reply With Quote

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