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

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#### Jameson

Staff member
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    * * * *   * * * *      |
|                                 |
* - - - - - - - - - - - - - - - - *
$$\text{Given: }\(A \cap B') \:=\:0.3$$

. . . . . . .$$P(A \cup B)' \:=\:0.4 \:=\(A' \cap B')$$

$$\text{Therefore: }\(B) \:=\:1 - 0.3 - 0.4 \:=\:0.3$$

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