Question

The following table shows the frequency of care for some animal species in a center specializing in veterinary dentistry. Species % Dog 52.8 Cat 19.2 Chinchilla 14.4 Marmoset 6.2 Consider that the center only serves 10 animals per week. For a given week, what is the probability that at least two are not dogs? ATTENTION: Provide the answer to exactly FOUR decimal places

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Andrea

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42 Answers

probability of atleast 1 are not dog
p=10C1(0.472)^1 (0.528)^0=0.015054
probability that all are dogs
p=10C0(0.472)^0(0.528)^10 =0.001684
1-0.015054-0.001684 = 0.983262
answer:
98.3262%

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