Use the probability distribution to find probabilities in parts (a) through (c). The probability distribution of num…
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Question “Use the probability distribution to find probabilities in parts (a) through (c). The probability distribution of num…”
Use the probability distribution to find probabilities in parts
(a) through (c).
The probability distribution of number of dogs per household in
a small town
Dogs 0 1 2 3 4 5
Households 0.680 0.191 0.079 0.029 0.0130 0. 008
(a) Find the probability of randomly selecting a household that
has fewer than two dogs.
0.871 (Round to three decimal places as needed.)
(b) Find the probability of randomly selecting a household that
has at least one dog.
0.320 (Round to three decimal
places as needed.)
(c) Find the probability of
randomly selecting a household that has between one and three
dogs, inclusive.
0.299 (Round to three decimal
places as needed.)
Answer
Probability of selecting randomly a household with fewer than 2 dogs
Probability of household having 2 dogs = probability that 0 dogs are present + probability of one dog = 0.680 + 0.91
= 0.871
The probability of randomly choosing a household with fewer than 2 dogs = 0.871
(b) Probability to randomly select a household with at least one dog
Probability of a household having at least one dog = probability that the household has more than 0 dogs
The probability that a household has zero dogs is known. The probability that a household has MORE dogs than 0 is the same as “all possible except 0 dogs”.
(“The probability that something is false = 1 – The probability that something is true”) This is known as a complement.
This is equal to = 1 – 0.6880
= 0.320
Chance of selecting a household with at least one dog = 0.320
(c) Probability to randomly select a household with between one and three dogs inclusive
Probability of 1-3 Dogs, Inclusive = probability 1 + probability 2 + probability 3
= 0.191 + 0.079 + 0.029
= 0.299
Probability to randomly select a household with one or three dogs inclusive = 0.299
Conclusion
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