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  1. #1
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    help help

    16 show that : p(A)+P(B)-1 ≤ P(A∩B) ≤ P(A)+P(B).
    17. If A and B are disjoint P(A) =0.37 , P(B)= 0.44 , compute :
    (a) P (A)=0.37
    (b) P (B^c)=1-0.44=
    (c) P(AUB)=
    (d)P(A∩B)=
    (e)P(A∩B^c)=
    (f) p (A^c∩B^c)=

    18. if P(A) =0.59 ,P(B)= 0.3 , P(A∩B)=0.21 compute :
    P (AUB)=
    P (A∩B^c)=
    p (A^cUB^c)=
    p (A^c∩B^c)=

    19 . show that the conditional probability satisfies the following
    if P(B)>0 , then
    i. P(A|B)≥ 0 ,
    ii. P(B|B)=1 ,
    iii.P[ A|B]=∑_(i=1)^∞▒〖p(A|B)〗
    for any sequence of disjoint events A1,A2,………..
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  2. #2
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    Hello, compufatwa!

    Here's the second one . . .


    17. If $\displaystyle A$ and $\displaystyle B$ are disjoint, and $\displaystyle P(A) = 0.37,\;P(B) = 0.44$, compute:
    $\displaystyle (a)\;P(A)\;=\;0.37$

    $\displaystyle (b)\;P(B^c)\;=\;1-0.44\;=\;0.56$

    $\displaystyle (c)\;P(A \cup B)\;=\;P(A) + P(B) - P(A \cap B) \;=\;0.37 + 0.44 - 0 \;=\;0.81$

    $\displaystyle (d)\;P(A \cap B)\;=\;0$

    $\displaystyle (e)\;P(A \cap B^c)\;=\;P(A) \;=\;0.38$

    $\displaystyle (f)\;P(A^c \cap B^c)\;= \;P(A \cup B)^c \;=\;1-0.81\;=\;0.19$

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  3. #3
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    In any probability space for every event X, $\displaystyle 0 \le P(X) \le 1$.
    So $\displaystyle 0 \le P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) \le 1\quad \Rightarrow P(X) + P(Y) - 1 \le P(X \cap Y)$.

    And $\displaystyle 0 \le P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)\quad \Rightarrow \quad P(X \cap Y) \le P(X) + P(Y)$.
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by compufatwa View Post
    18. if P(A) =0.59 ,P(B)= 0.3 , P(A∩B)=0.21 compute :
    P (AUB)=
    P (A∩B^c)=
    p (A^cUB^c)=
    p (A^c∩B^c)=
    $\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.59 + 0.3 - 0.21$
    for the second, i haven't solved yet.. Ü, but i think, it's just the same with the previous post..
    or $\displaystyle P(A \cap B^c) = P(A) - P(A \cap B) $ .... not sure..
    $\displaystyle P(A^c \cup B^c) = P(A \cap B)^c = 1 - P(A \cap B)$
    same process with the fourth, apply de morgan's law..

    Quote Originally Posted by compufatwa View Post
    19 . show that the conditional probability satisfies the following
    if P(B)>0 , then
    i. P(A|B)≥ 0 ,
    ii. P(B|B)=1 ,
    iii.P[ A|B]=∑_(i=1)^∞▒〖p(A|B)〗
    for any sequence of disjoint events A1,A2,………..
    $\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)}$
    but since $\displaystyle P(\cdot) \geq 0$ and P(B)>0,
    therefore $\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)} \geq 0$

    $\displaystyle P(B|B) = \frac{P(B \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1$
    i dont understand the code for the third one..
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