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Math Help - home assignment

  1. #1
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    home assignment


    Attention please!!!!!!!


    a. In which situation, we use Permutation and Combination.

    b. An MBA applies for a job in two firms X and Y. The probability of being selected in firm X is 0.7 and being rejected at Y is 0.5. The probability of at least one of his applications being rejected is 0.6. What is the probability that he will be selected by one of the firm?

    c. Husband and wife appear in an interview for two vacancies in the same post. The probability of husbandís selection is and that of wife is. What is the probability that:

    1. Both of them will selected
    2. Only one of them is selected
    3. None of them is selected.
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  2. #2
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    Hello, MS BATOOL!

    b) An MBA applies for a job in two firms X and Y.
    The probability of being selected in firm X is 0.7 and being rejected at Y is 0.5.
    The probability of at least one of his applications being rejected is 0.6.
    What is the probability that he will be selected by one of the firm?
    We want: . P (X \vee Y)


    We are given: . \begin{array}{ccccccc}P(X) &=& 0.7 &\quad & P(X') & = & 0.3\\<br />
P(Y) & = & 0.5 & \quad & P(Y') & = & 0.5 \end{array}\qquad P(X' \vee Y') \:=\:0.6

    \text{Then: }\;\underbrace{P(X' \vee Y')}\;=\;\underbrace{P(X')} + \underbrace{P(Y')} - P(X' \wedge Y')
    Hence: . . 0.6 \qquad\,=\quad 0.3 \;\;\,+\;\; 0.5 - \;P(X' \wedge Y')\qquad\Rightarrow\qquad P(X' \wedge Y') \:=\:0.2


    By DeMorgan's Law: . P(X \vee Y) \;=\;P(X' \wedge Y')' \;=\;(0.2)' \;=\;\boxed{0.8}



    c) Husband and wife appear in an interview for two vacancies in the same post.
    The probability of husbandís selection is ?? and that of wife is ??

    Obviously, we need more information.
    Also, can we assume that the events are independent?


    What is the probability that:

    1. Both of them will selected
    2. Only one of them is selected
    3. Neither of them is selected.
    Let: . \begin{array}{ccc}P(H) \:=\:h & \quad & P(H') \:=\:1-h \\ P(W) \:=\:w & \quad & P(W') \:=\:1-w\end{array}


    1)\;P(\text{Both selected}) \;=\;P(H)\cdot P(W) \;=\;h\cdot w

    2)\;P(\text{One selected}) \;=\;P(H)\cdot P(W') + (H')\cdot P(W) \;=\;h(1-w) + (1-h)w

    3)\;P(\text{Neither selected}) \;=\;P(H')\cdot P(W') \;=\;(1-h)(1-w)

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  3. #3
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    thanks

    Respected Sir,
    i hope you will be fine.......yeah we can assume that the events are independent........i just want to thank you for your usefull reply.
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  4. #4
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    Question

    hello Sir !
    i want to ask you ....can you help me in other threads about Statistics which are sent by me.....actually today is the last day to submit the assignments......i will be very thankful to you for this act of kindness.
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