1. Upon arrival at a hospital’s emergency room, patients are categorized according to their conditions as critical, serious, or stable. In the past year, 10% were classified as critical, 30% serious, and the rest were stable. The records of 12 emergency room patients are randomly chosen to be examined for analysis. What is the probability the first critical patient record was the fourth one examined?

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1. Upon arrival at a hospital’s emergency room, patients are categorized according to their conditions as critical, serious, or stable. In the past year, 10% were classified as critical, 30% serious, and the rest were stable. The records of 12 emergency room patients are randomly chosen to be examined for analysis. What is the probability the first critical patient record was the fourth one examined?

Mathematics
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It really depends on how many patients they had in the past year to get a good answer. However I think they are assuming it says at 10% even the 1st, 2nd, and 3rd patients are selected. Therefore it would be .9 * .9 * .9 * .1 = 7.29%
You multiple the odds of the first 3 of it not being a critical patient, since that is what it asks for. Then it asks for the fourth to be critical. 90% chance for it to not be critical and 10% for it to be critical. Just multiply the odds.
It actually does not depend on how may they had last year. I was thinking of something else.

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it actually doesn´t depend on how many pacients they had las year, by saying that 10 per cent are critical, they say that for every 100 patients 10 are critical

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