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no rain monday ; .6 and for tuesday ; .2
atleast rain on 1 of the days ; 1-(.6*.2) ; 1-.12 ; .88
or say 88%
IMO B

Bunuel
The chance of rain in Dennyville is 40% on Monday and 80% on Tuesday. What is the probability that it rains in Dennyville on at least one of these two days?

A. 120%
B. 88%
C. 76%
D. 60%
E. 32%
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Bunuel
The chance of rain in Dennyville is 40% on Monday and 80% on Tuesday. What is the probability that it rains in Dennyville on at least one of these two days?

A. 120%
B. 88%
C. 76%
D. 60%
E. 32%

The probability that it rains on at least one day is:

P (rain at least one day) = 1 - (probability of no rain on each day)

P (rain at least one day) = 1 - 0.6 x 0.2

P (rain at least one day) = 1 - 0.12 = 0.88

Answer: B
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Bunuel
The chance of rain in Dennyville is 40% on Monday and 80% on Tuesday. What is the probability that it rains in Dennyville on at least one of these two days?

A. 120%
B. 88%
C. 76%
D. 60%
E. 32%

We want P(at least one day of rain)
When it comes to probability questions involving "at least," it's usually best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(at least one day of rain) = 1 - P(not getting at least one day of rain)
What does it mean to not get at least one day of rain? It means getting zero rain.
So, we can write: P(getting at least one day of rain) = 1 - P(getting zero days of rain)

P(getting zero days of rain)
P(getting zero days of rain) = P(no rain on Monday AND no rain on Tuesday)
= P(no rain on Monday) x P(no rain on Tuesday)

Given: P(rain on Monday) = 40% = 0.4, which means = P(NO rain on Monday) = 60% = 0.6
Given: P(rain on Tuesday) = 80% = 0.8, which means = P(NO rain on Tuesday) = 20% = 0.2

So, P(getting zero days of rain) = 0.6 x 0.2
= 0.12

So, P(getting at least one day of rain) = 1 - 0.12
= 0.88
= 88%

Answer: B

Cheers,
Brent
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