Titleist wrote:
This is what I get so far using the binomial theorem:
rain 5 days in a row 7C5*(3/4)^5*(1/4)^2
rain 6 days in a row7C6*(3/4)^6*(1/4)^1
rain 7 days in a row7C7*(3/4)^7*(1/4)^0
Add em up and you get
(3/4)^5*(1/4)^2*(7!/5!2!)+(3/4)^6*(1/4)*(7!/6!)+(3/4)^7which is
E in
MGMAT's Answer choice
do i get underpants just for trying???
what did you get duttsit?
Looks like we need to relook at this question. There is more than what meets the eyes.
First of all, using theorem, we can find that:
p(rain) = 3/4
p(no rain) = 1/4
we need to find:
p(rain for at least 5 days in a row)
this is equivalent to:
p(raining 5 days in a row) + p(raining 6 days in a row,
5 of which should be in a row) + p(raining for all 7 days)
note: second component, it is not 6 days in a row
R: Rain
N: No rain
lets find invidual components:
p(raining 5 days in a row) =
RRRRRNN
NRRRRRN
NNRRRRR
each of these have probability: (3/4)^5 (1/4)^2
so p(raining 5 days in a row) = 3 * (3/4)^5 (1/4)^2
p(raining 6 days in a row,
5 of which should be in a row) =
RRRRRRN
NRRRRRR
RRRRRNR
RNRRRRR
each of these have probability (3/4)^6 * (1/4)^1
p(raining 6 days in a row,
5 of which should be in a row) =
4 * (3/4)^6 * (1/4)
p(raining for all 7 days) = (3/4)^7 simple
plugging all these components back gives:
p(raining atleast 5 days in a row) =
3 * (3/4)^5 (1/4)^2 + 4 * (3/4)^6 * (1/4)
+ (3/4)^7