Venn Diagrams and Probabilities

Here's my problem:

Use a Venn Diagram to work this exercise:

P(z) = .40
P(y) = .30
P(zuy) = .58

Find each of the given probabilities:

P(z' upside down u y')

p(z'uy')

p(z'uy)

p(z upside down u y')

I'm pretty sure that I think I know that .40 would be the total in the z circle and .30 would be the total in the y circle but beyond that I don't have a clue. I'm really cofused over what each of these terms mean and how I would figure them out in a venn diagram. Any help would be appreciated!

Hello, lizzpalmer!

\(\displaystyle \text{Given: }\:\begin{Bmatrix} {(Z) &=& 0.40 \\ P(Y) &=& 0.30 \\ P(Z \cup Y) &_& 0.58 \end{Bmatrix}\)

\(\displaystyle \text{Use a Venn diagram to find these probabilities:}\)

\(\displaystyle (a)\;P(Z' \cap Y') \qquad (b)\;P(Z' \cup Y') \qquad (c)\;P(Z' \cup Y) \qquad (d)\;P(Z \cap Y')\;\)

Click to expand...

With only two sets, we can set up a table.

. . \(\displaystyle \begin{array}{|c||c|c||c|}\hline & Y & Y' & \text{Total} \\ \hline\hline Z & & & \\ \hline Z' & & & \\ \hline\hline\text{Total} & & & 1.00 \\ \hline \end{array}\)


\(\displaystyle \text{Formula: }\(Z \cup Y) \:=\: P(Z) + P(Y) - P(Z \cap Y)\)

. . . . . . . . . . .\(\displaystyle 0.58 \;=\; 0.40 + 0.30 - P(Z \cap Y)\)

. . . . . . .\(\displaystyle P(Z \cap Y) \:=\:0.12\)


We can fill in the known data:

. . \(\displaystyle \begin{array}{|c||c|c||c|} \hline & Y & Y' & \text{Total} \\ \hline\hline Z & 0.12 & & 0.40 \\ \hline Z' & & & 0.60 \\ \hline\hline\text{Total} & 0.30 & 0.70 & 1.00 \\ \hline \end{array}\)


And we can fill in the rest of the cells:

. . \(\displaystyle \begin{array}{|c||c|c||c|} \hline & Y & Y' & \text{Total} \\ \hline\hline Z & 0.12 & 0.28 & 0.40 \\ \hline Z' & 0.18 & 0.42 & 0.60 \\ \hline\hline\text{Total} & 0.30 & 0.70 & 1.00 \\ \hline \end{array}\)


Now you should be able to answer the questions.

You have no idea how much that helped me. I had no idea you could do it in a table!

lizzpalmer said:

You have no idea how much that helped me. I had no idea you could do it in a table!

Click to expand...

In one regard, all of math is a HUGE bag of tricks for making simple what appears to be insoluble.