In: Statistics and Probability
1) The range of probability is
c) zero to one.
2) Any process that generates well-defined outcomes is
b) an experiment
3) The sample space refers to
c) the set of all possible experimental outcomes
4) An experiment consists of three steps. There are four possible results on the first step, three possible results on the second step, and two possible results on the third step. The total number of experimental outcomes is
4*3*2 = 24
c) 24
5) When the assumption of equally likely outcomes is used to assign probability values, the method used to assign probabilities is referred to as the
d. classical method
6) Given that event E has a probability of 0.25, the probability of the complement of event E
P(Ec ) = 1- P(E) = 1-0.25 = 0.75
c) must be 0.75
7)If P(A) = 0.38, P(B) = 0.83, and P(A ∩ B) = 0.57; then P(A ∪ B) =
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
= 0.38+0.83-0.57 = 0.64
b) 0.64
14) If P(A) = 0.62, P(B) = 0.47, and P(A ∪ B) = 0.88; then P(A ∩ B) =
P(A ∩ B) = P(A) + P(B) - P(A ∪ B)
= 0.62+0.47-0.88 = 0.21
d) 0.2100
15)Two events are mutually exclusive if
b) they have no sample points in common
16) If A and B are mutually exclusive events with P(A) = 0.3 and P(B) = 0.5, then P(A ∪ B) =
P(A ∪ B) = P(A) + P(B) = 0.3+0.5 = 0.8
c) 0.8
17) If P(A|B) = P(A) then
d)A and B are independent.
18) On a December day, the probability of snow is .30. The probability of a "cold" day is .50. The probability of snow and a "cold" day is .15. Are snow and "cold" weather independent events
P(snow and cold) = 0.15 = P(snow) * P(cold) = 0.3*0.5 = 0.15
c) yes
19) If A and B are independent events with P(A) = 0.4 and P(B) = 0.6, then P(A ∩ B) =
P(A ∩ B) = P(A) * P(B) = 0.4*0.6 = 0.24
c) 0.24
20) If A and B are independent events with P(A) = 0.2 and P(B) = 0.6, then P(A ∪ B) =
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
= P(A) + P(B) - P(A)*P(B)
= 0.2+0.6 - 0.2*0.6 = 0.68
d) 0.68
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