let us take this problem as if we are asked to place 3 R (rain days) and 4 Nr(non rain days) in 7 place

condition all R has to be togather

__1 __2 __3 __4 ___5 __6__7

now R can take any value from 1 to 5 ( in order to be consecutive)
it can be done in 5C1 ways=5
now 4Nr can be arranged in 4 left places in one way ( as order does not matters)

tot ways=5
now probabilty of a rain=.3
hence for 3 rain days P=.3*.3*.3
and for 4 Nr P=.7*.7*.7*.7

fnal P=5*(.3^)*(.7^4)=.0324

coscha wrote:If the probability that it will rain on any day in September is 0.3, caluclate the probability that in a given week in September, it will rain on exactly three consecutive days.

I got 0.027...but the answer is supposed to be 0.0324

Help plz

The rain can start with first day (1st day 2nd day and 3rd day rains) Second day , third day or fourth day and 5th day

In case the rain starts with first day it has to rain first day second and third day, additionally it should not rain on 4th 5th 6th and 7th days ; probability = 1/3*1/3*1/3*2/3*2/3*2/3*2/3 = 16 /3^6

similarly probability for rains starting on second day is = 16/3^6... and so on so forth till 5th day

total probability = 80/3^6

Hey ajith, 0.3 = 3/10 and not 1/3..and probability of not raining is 7/10 not 2/3 ? I guess you are tired

ace_gre wrote:Hey ajith, 0.3 = 3/10 and not 1/3..and probability of not raining is 7/10 not 2/3 ? I guess you are tired

Ahhh I get it... I guess i am tired too, Thanks

Thanks alot guys