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Oct 2, 2020 at 3:19 vote accept user13232774
Sep 29, 2020 at 9:00 history tweeted twitter.com/StackStats/status/1310866925476601856
Sep 29, 2020 at 4:46 comment added user13232774 @BruceET Yes, that assumption is correct
Sep 29, 2020 at 4:34 answer added BruceET timeline score: 2
Sep 29, 2020 at 4:18 comment added BruceET Am I correct that 23 other guests had a chance to draw a number above 95? Then the probability of a higher number is $1 - {94 \choose 23}/{98 \choose 23} = 0.6635228.$ Or in R, where dhyper is a hypergeometric PDF, 1-dhyper(0, 4, 94, 23) returns $0.6635228.$
Sep 29, 2020 at 3:53 review First posts
Sep 29, 2020 at 4:19
Sep 29, 2020 at 3:52 comment added Dave I guess you win by drawing the highest number?
Sep 29, 2020 at 3:48 history asked user13232774 CC BY-SA 4.0