Question

There is a 20 percent chance of rain today and a 80 percent chance of rain tomorrow. If the two events are independent, what is the probability that it will rain today or tomorrow?

A. 28
B. 72
C. 18
D. 84
E. 100

Answer 1

Hmm... it's definitely not E, I can give you that. And it can't be A or C either.

Answer 2

d

Answer 3

Actually, I lied about A/C. I can't say for sure that those are wrong, but I mean I KNOW logically it's not E, and it seems as though A and C won't work because they're both lower than 20.

Answer 4

zarkon help me answer atleast one after you done please

Answer 5

surely the answer is 0.68

Answer 6

P(rain today and not tomorrow) = 0.2 * 0.2 = .04
P(no rain today and rain tomorrow) = 0.8 * 0.8 = 0.64
p of one or the other = sum 0.04 + 0.64 = 0.68 or 68%

Answer 7

it is 84%

Answer 8

84 per cent is probability that it wont rain on both days 1-0.16 = 0.84
0.16 is probability it will rain on both days

Answer 9

let
\[R_1\] be rain today
and \[R_2\] be rain tomorrow
we want

\[P(R_1\cup R_2)=P(R_1)+P(R_2)-P(R_1\cap R_2)=P(R_1)+P(R_2)-P(R_1)\cdot P(R_2)\]
=.2+.8-.2*.8=.84

Answer 10

note: for any two events

\[P(A\text{ or }B)\geq P(A)\]

and
\[P(A\text{ or }B)\geq P(B)\]

therefore the answer has to be larger that 80%

Answer 11

i bow to your superior knowledge of set theory applied to probability, but i really cant see what is wrong with my argument. i'll have to check it out.

Answer 12

if you wanted the probability it would not rain on both days then that would be

\[P(R_1^{c}\text{ and }R_2^{c})\]

by independence

\[=P(R_1^{c})\cdot P(R_2^{c})=(1-.2)\cdot(1-.8)=.8\cdot .2=.16\]

therfore the

\[P(R_1\cup R_2)=1-P(R_1^{c}\text{ and }R_2^{c})=1-.16=.84\]