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Problem of the week #7 - May 14th, 2012

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Jameson

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Jan 26, 2012
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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) Sudharaka
2) checkittwice
3) soroban

Solution:

We are given that \(\displaystyle P[(A \cup B)']=0.4\), which is equivalent to saying that \(\displaystyle P[A' \cap B']=0.4\). Using the identity \(\displaystyle P[B']=P[A\cap B']+P[A' \cap B']\) we can solve for P[B'], which is 0.7. Finally we can solve for P by \(\displaystyle P=1-P[B']=1-0.7=0.3\). So we have a final answer of 0.3


Here is a similar solution from MHB member soroban with a nice diagram:


Draw a Venn diagram . . .

Code:
      * - - - - - - - - - - - - - - - - *
      |                                 | 
      |          * * * *   * * * *      | 
      |         *       * *.......*     |
      |        *    A    *....B....*    | 
      |       *         *.*.........*   |
      |      *         *...*.........*  | 
      |      *         *...*.........*  | 
      |      *   0.3   *...*.........*  | 
      |      *         *...*.........*  |
      |      *         *...*.........*  | 
      |       *         *.*.........*   | 
      |        *         *.........*    | 
      |         *       * *.......*     | 
      |   0.4    * * * *   * * * *      | 
      |                                 |
      * - - - - - - - - - - - - - - - - *
[tex]\text{Given: }\:p(A \cap B') \:=\:0.3 [/tex]

. . . . . . .[tex]P(A \cup B)' \:=\:0.4 \:=\:p(A' \cap B')[/tex]


[tex]\text{Therefore: }\:p(B) \:=\:1 - 0.3 - 0.4 \:=\:0.3[/tex]

 
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