Assignment 5 Of MTH202 (Spring 2010)

[Death race]

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.


plz solve the assignments

[Death race]

last date is 19 plz solve it

[hanif]

plz koi idea provide kar dy

[Xpert]

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

[viki]

please make the discussion and solve it

[Xpert]

yes thats the right way to solve it. share ur knowladge

[zoob2010]

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.

[Xpert]

So what is the conclusion ?

[Death race]

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

[viki]

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

[adeel_99f]

thanks
dear

[Xpert]

you are welcome dear.