Question

can any one help me through this?
the 5 day forecast says there will be a 70% chance of rain each day the next 5 days in Tuxtown. The Tuxtown soccer team needs to schedule their practices for the week, but they cant practice if it rains. Using (p+q)^5 as a model, what is the probability they will practice 3 days and be rained out 2 days? what is the probability they can practice all 5 days?

Answer 1

what does the (p+q)^5 do for us?

Answer 2

the expanded from defines the individual probabilities:

P(rain 0 days) + P(rain 1 day) + P(rain 2 days) + P(rain 3 days)
+ P(rain 4 days) + P(rain 5 days)

now, a probability distribution is only valid when the sum of all events is equal to 1, therefore: (p+q)^5 = 1 only if p+q = 1

if p is the probability of rain, then q is the probability of no rain.

Answer 3

i dont quite understand, would i only need the p(rain 2 days) +p(rain 0 days)?

Answer 4

no, you are not adding them

P(rain 2 days) means that they practice for 3 days, and get rained out 2 days.

P(rain 0 days) means that they practice for 5 days, and get rained out 0 days.

the only time we add them up, is if we want a cumulative probability, say whats the probability that they practice at most 3 days,

P(rains 2) + P(rains 3) + P(rains 4) + P(rains 5) = 1 - P(rains 0) + P(rains 2)

Answer 5

(sorry late reply, i fell asleep)
so im just trying to find the two different probability's then separately?
then the first one would be

Answer 6

lets expand the (p+q)^5 and let p = .30 (no rain), q=.70 (rain)
\[\underbrace{1p^5q^0}_{P(0)}
+ \underbrace{5p^4q^1}_{P(1)}
+\underbrace{10p^3q^2}_{P(2)}
+\underbrace{10p^2q^3}_{P(3)}
+\underbrace{5p^1q^4}_{P(4)}
+\underbrace{1p^0q^5}_{P(5)}
\]