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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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Answer to a math 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

Expert avatar
Andrea
4.5
73 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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