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At a local clinic there are 8 men, 5 women, and 3 children in the waiting room. If3 patients are randomly selected,?

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  • At a local clinic there are 8 men, 5 women, and 3 children in the waiting room. If3 patients are randomly selected,?


Answers

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Answer #2 | 15/05 2016 20:14
Let's find the opposite probability... that *no* children are selected. First from the 13 adults (8 men and 5 women), choose 3 patients. C(13,3) = (13 x 12 x 11) / (3 x 2 x 1) = 286 ways Then from the 16 total patients, select *any* 3. C(16,3) = (16 x 15 x 14) / (3 x 2 x 1) = 560 ways Divide to get the probability of only picking adults (NO children) = 286/560 = 143/280 But you want the opposite probability (at least one child), so subtract from 1. P(at least one child) = 1 - 143/280 = 137/280
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Answer #9 | 15/05 2016 13:43
use the compliment at least one child is the same as 1 - no child there are 13 people that aren't children there are 16 people total the probability of no children selected is 13C3 / 16C3 so the probability that at least one child is selected is 1 - (13C3 / 16C3) (16C3 - 13C3) / 16C3 (560 - 286) / 560 274 / 560 137 / 280
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Answer #20 | 15/05 2016 20:43
use the compliment at least one child is the same as 1 - no child there are 13 people that aren't children there are 16 people total the probability of no children selected is 13C3 / 16C3 so the probability that at least one child is selected is 1 - (13C3 / 16C3) (16C3 - 13C3) / 16C3 (560 - 286) / 560 274 / 560 137 / 280
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Answer #42 | 15/05 2016 13:14
Let's find the opposite probability... that *no* children are selected. First from the 13 adults (8 men and 5 women), choose 3 patients. C(13,3) = (13 x 12 x 11) / (3 x 2 x 1) = 286 ways Then from the 16 total patients, select *any* 3. C(16,3) = (16 x 15 x 14) / (3 x 2 x 1) = 560 ways Divide to get the probability of only picking adults (NO children) = 286/560 = 143/280 But you want the opposite probability (at least one child), so subtract from 1. P(at least one child) = 1 - 143/280 = 137/280
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