1. The probability of a dry summer is equal to 0.3, the probability of a wet sum
ID: 3060385 • Letter: 1
Question
1. The probability of a dry summer is equal to 0.3, the probability of a wet summer is equal to 0.2, and the probability of a summer with normal precipitation is equal to 0.5. A climatologist observed the precipitation during three consecutive summers. (a) How many outcomes are in the sample space? Enumerate the sample space, and assign probabilities to each simple event. (b) What is the probability of observing two dry summers? (c) What is the probability of observing at least two dry summers? (d) What is the probability of not observing a wet summer? 1. The probability of a dry summer is equal to 0.3, the probability of a wet summer is equal to 0.2, and the probability of a summer with normal precipitation is equal to 0.5. A climatologist observed the precipitation during three consecutive summers. (a) How many outcomes are in the sample space? Enumerate the sample space, and assign probabilities to each simple event. (b) What is the probability of observing two dry summers? (c) What is the probability of observing at least two dry summers? (d) What is the probability of not observing a wet summer? 1. The probability of a dry summer is equal to 0.3, the probability of a wet summer is equal to 0.2, and the probability of a summer with normal precipitation is equal to 0.5. A climatologist observed the precipitation during three consecutive summers. (a) How many outcomes are in the sample space? Enumerate the sample space, and assign probabilities to each simple event. (b) What is the probability of observing two dry summers? (c) What is the probability of observing at least two dry summers? (d) What is the probability of not observing a wet summer?Explanation / Answer
a)Sample space:
b)
P(observing two dry summers) = 0.3*0.3 = 0.09
c)P(observing at least two dry summers)
=> 2 dry summers or 3 dry summers
= 0.3*0.3 + 0.3*0.3*0.3 = 0.117
d)
P(not observing a wet summer) = 1 - ( P(observing 1 wet summer) + P(observing 2 wet summer) + P(observing 3 wet summers) )
= 1 - ( 0.2 + 0.2*0.2 + 0.2*0.2*0.2)
= 1 - 0.248
= 0.752
Type of summer(X) Dry Wet Normal Probability(P) 0.3 0.2 0.5Related Questions
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