Question

Each week Van and Marie take part in a raffle at their respective workplaces. the probability that Van wins a prize in his raffle is 1/9 and Marie, 1/16. What is the probability that during the next 3 weeks, at least one of them wins a prize?

Answer 1

is the answer 107/218 ???

Answer 2

if Vwin=1/9 then Vlost=8/9
if Mwin=1/16 then Mlost=15/16
divided 3 chase :
* P(Vwin and Mlost) = 1/9 * 15/16
** P(Vlost and Mwin) = 8/9 * 1/16
*** P(Vwin and Mwin) = 1/9 * 1/16
total = (15+8+1)/(9*16) = 24/144
because during 3 week, it can be (24/144)*3 = 72/144 = 1/2

Answer 3

@bhaskarbabu, the answer is 91/216
and @RadEn, I understood how you got it but it's not the right answer. I'm so confused!

Answer 4

I am going out now...bhaskarbabu is my friend and i will meet him. He or I will be back with a correct solution, I promise.

Answer 5

probably easier to compute the probability that neither wins, subtract from one
\[(\frac{8}{9})^2\] for one and
\[(\frac{15}{16})^3\] for the other, together it would be
\[(\frac{8}{9})^3\times (\frac{15}{16})^3\] your answer is therefore
\[1-(\frac{8}{9})^3(\frac{15}{16})^3\]

Answer 6

yea, @satellite73 was easier method, u have used complement probably right, satellite ?

Answer 7

@satellite73 why do you multiply the two probabilities of neither of them winning instead of adding? and @RadEn, what was it that you found before?

Answer 8

we knowed that if given the probility of an even, then the probability its complement can be found substracting the given probability from 1.

complement of win is lost, so
P(V lost) = 1-(1/9)=8/9
P(M lost) = 1-(1/16)=15/16
and probability complement even of "at least one of them wins a prize" similar with probability "Van and Marie always lost"!
so, P(V lost during 3 weeks) = 8/9 * 8/9 * 8/9 = (8/9)^3
P(M lost during 3 weeks ) = 15/16 * 15/16 * 15/16 = (15/16)^3
so, probility of Van and Marie always lost in 3 weeks is :