bblast wrote:the easiest approach in at least probablility is
1-non favourable outcomes :
what is non favourable ?
1>it rains exactly 1 day - > P=(1/2)
2>it rains exactly 2 days- > P=(1/2*1/2)
add both we get 1/2+1/4 = 3/4
hence the answer follows :
1-3/4 = 1/4
cheers !!
Be careful, bblast. You were lucky.
For many problems, the following strategy works well:
P(good outcomes) = 1 - P(bad outcomes).
But your reasoning overlooks quite a few bad outcomes, bolded in red below:
P(at least 3 days of rain in a ROW) = 1 - P(no rain) - P(exactly 1 day of rain) - P(exactly 2 days of rain) - P(exactly 3 days of rain NOT in a row) - P(exactly 4 days of rain without 3 days of rain in a row)
Your calculations are not quite correct. If R=rain and N=no rain:
P(exactly 1 R):
P(RNNNN) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32.
Since R can occur on any of the 5 days, we must multiply the result above by 5:
P(exactly 1 R) = 5*(1/32) = 5/32.
P(exactly 2 R):
P(RRNNN) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32.
The number of ways to arrange RRNNN = 5!/(2!3!) = 10. Thus, to account for all the ways in which we could get two R's, the result above must be multiplied by 10.
P(exactly 2 days of rain) = 10*(1/32) = 10/32.
Subtracting these fractions from 1, we get:
1 - 5/32 - 10/32 = 17/32, which is not the correct answer.
So please be careful. Your approach will work well for many problems, but it's important to count ALL the different ways a bad outcome could occur.
In this particular problem, I think it's easier to determine P(at least 3 R in a row) directly:
Total possible outcomes:
Each day could be R or N, giving us 2 possible outcomes for each day.
Over 5 days, total possible outcomes = 2*2*2*2*2 = 32.
Good outcomes:
Exactly 3 days of rain in a row:
RRRNN
RRRNR
NRRRN
NNRRR
RNRRR
5 good outcomes.
Exactly 4 days of rain in a row:
RRRRN
NRRRR
2 good outcomes.
Exactly 5 days of rain in a row:
RRRRR
1 good outcome.
Good outcomes = 5+2+1 = 8.
Thus, P(at least 3 days of rain in a row) = 8/32 = 1/4.
The correct answer is
B.
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