Answer to $P[B]$ in $P[A\cap B']=0.3$ and $P[(A \cup B)']=0.4$

[Jameson]

$P[A\cap B']=0.3$ and $P[(A \cup B)']=0.4$ . What is $P$?

 

[Jameson]

Congratulations to the following members for their correct solutions:

1) Sudharaka
2) checkittwice
3) soroban

Solution:

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

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

[sp]
Draw a Venn diagram . . .

      * - - - - - - - - - - - - - - - - *
      |                                 | 
      |          * * * *   * * * *      | 
      |         *       * *...*     |
      |        *    A    *...B...*    | 
      |       *         *.*...*   |
      |      *         *...*...*  | 
      |      *         *...*...*  | 
      |      *   0.3   *...*...*  | 
      |      *         *...*...*  |
      |      *         *...*...*  | 
      |       *         *.*...*   | 
      |        *         *...*    | 
      |         *       * *...*     | 
      |   0.4    * * * *   * * * *      | 
      |                                 |
      * - - - - - - - - - - - - - - - - *

$$\text{Given: }\:P(A \cap B') \:=\:0.3$$

. . . . . . .[/color]$$P(A \cup B)' \:=\:0.4 \:=\:P(A' \cap B')$$$$\text{Therefore: }\:P(B) \:=\:1 - 0.3 - 0.4 \:=\:0.3$$

[/size] [/sp]