Stat Question about independent events

Knowing the values of P(a) and P(b), how would would you calculate the following if a and b are independent events:

1.)P(a^b')
2.)P(a'^b')
3.)P[(avb)' ]
4.)P(a'^b)

^ = intersection
v = union

I know that P(a^b)=P(a)P(b), but i have no clue of the formula to achieve the above. Someone providing the formula or a good online resource that answers these would be helpful. Thanks.

Hwllo, dhs316!

Knowing the values of \(\displaystyle P(a)\) and \(\displaystyle P(b)\),
and knowing that \(\displaystyle a\) and \(\displaystyle b\) are independent events,
calculate the following:

\(\displaystyle (1)\;\;P(a \cap b ')\qquad (2)\;\;P(a' \cap b') \qquad (3)\;\;P(a\cup b)' \qquad (4)\;\;P(a' \cap b)\)

Click to expand...

You are right!

\(\displaystyle \text{Since }a\text{ and }b\text{ are independent, then: }\(a \cap b) \:=\(a)\cdot P(b)\)

\(\displaystyle \text{We also know the values of: }\;\begin{array}{ccc}P(a') &=& 1 - P(a) \\ P(b') &=& 1 - P(b)\end{array}\)


\(\displaystyle (1)\;\;P(a \cap b') \;=\;P(a) \cdot P(b') \;=\;P(a)\cdot\left[1-P(b)\right]\)


\(\displaystyle (2)\;\;P(a'\cap b') \;=\;P(a')\cdot P(b') \;=\;\left[1-P(a)\right]\cdot\left[1-P(b)\right]\)


\(\displaystyle (3)\;\;P(a \cup b)' \;=\;P(a' \cap b')\:\text{ by DeMorgan's Law}\)
. . . \(\displaystyle \text{This is the same as (2)}\)


\(\displaystyle (4)\;\;P(a' \cap b) \;=\;P(a')\cdot P(b) \;=\;[1-P(a)]\cdot P(b)\)

Thank you. i never looked at solving these problems in the way that you did.