Math-Law-other-studies: Testing Blogger LaTeX with a little more care. "Work with me now"

Saturday, January 13, 2007

Testing Blogger LaTeX with a little more care. "Work with me now"

Problem 1.2-6 (c)

Find $P(A' \bigcup B')$

Again using basic properties and theorems of probability

$P(A') =1-P(A) = 1-0.4 =0.6$

and $P(B') = 1-P(B) = 1- 0.5 = 0.5$

Now we need to use this formula

$P(A' \bigcup B') = P(A') + P(B') - P( A' \bigcap B')$ in this problem.

But $P (A \bigcup B) = 0.6$ and $P( (A \bigcup B)')$ its complement is simply

$1- P (A \bigcup B) = 1- 0.6 =0.4$

again by basic probability theorems

What is the set $(A \bigcup B)'$ actually? It is the area out side of both A and B. which is the same as an intersection of the complements of A and B or set $( A' \bigcap B')$

Therefore $P( A' \bigcap B') = 0.4$

Therefore substituting in our formula.

$P(A' \bigcup B') = 0.6 + 0.5 - 0.4 = 0.7$

And the set $(A' \bigcup B')$ is really the complement of $( A \bigcap B)$ which = 0.3 because outside of the intersection is outside of either A or B so all this space outside the intersection is in $(A' \bigcup B')$ . Thus we can check our work by checking that $P(A' \bigcup B')$ = 1- $P( A \bigcap B)$ = 1- 0.3 = 0.7