Stat Question about independent events

[dhs316]

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.

 

[soroban]

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:

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

You are right!

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

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


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


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


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


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

 

[dhs316]

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