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Thread: Assignment 5 Of MTH202 (Spring 2010)

  1. #1
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    Icon51 Assignment 5 Of MTH202 (Spring 2010)

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    Maximum Marks: 15
    Due Date: July 20, 2010

    DON’T MISS THESE: Important instructions before attempting the solution of this assignment:
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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

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    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
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  2. #2
    Junior Member Death race's Avatar
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    last date is 19 plz solve it

  3. #3
    Junior Member hanif's Avatar
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    plz koi idea provide kar dy

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

  5. #5
    Senior Member viki's Avatar
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    please make the discussion and solve it
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    yes thats the right way to solve it. share ur knowladge

  7. #7
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    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.

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    So what is the conclusion ?

  9. #9
    Junior Member Death race's Avatar
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    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

  10. #10
    Senior Member viki's Avatar
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    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
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