phoenixhazard wrote:kvcpk wrote:arora007 wrote:In Rwanda, the chance for rain on any given day is 50%. What is the probability that it rains on 4 out of 7 consecutive days in Rwanda?
a) 4/7
b) 3/7
c) 35/128
d) 4/28
e) 28/135
Good one....
4 days can be picked from 7 days in 7c4 ways = 35 possible ways.
On a single selection, probability of raining is 1/2 * 1/2 * 1/2 *1/2 * 1/2 * 1/2 * 1/2 = 1/(2^7)
So total = 35/128
I don't completelty:
7! / 4! * 3!
7 is total amount of options
4 is how many options u need in a group
3 is the remaining options
you get 35 total combinations that it will rain 4 days in any 7 of those days
then u need total chance
50% per day over 7 days
0.5 ^ 7
1/128
35 combinations * 1/128 probability it will rain all 7 days
.
is this the right logic?
Let R = rain, N = no rain.
P(RRRRNNN) = (1/2)^7 = 1/128.
We need to multiply this result by the number of ways to arrange RRRRNNN in order to account for ALL the different ways we could get 4 days of R.
The number of ways to arrange 7 elements is 7!. But when we have repeated elements such as RRRR and NNN, the number of unique arrangements will be reduced. To account for the four R's, we need to divide by 4!, and to account for the 3 N's, we need to divide by 3!: 7!/(4!*3!) = 35.
Thus, P(4 days of R) = 35*(1/128) = 35/128.
Does this help?
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