Help with Statistics (Supposedly Easy, But I'm LOST!) :(? - XP Math - Forums

 XP Math - Forums Help with Statistics (Supposedly Easy, But I'm LOST!) :(?

 07-25-2007 #1 math hobbyist Guest   Posts: n/a Help with Statistics (Supposedly Easy, But I'm LOST!) :(? I am having an impossible time with my statistics course. It's fully online, the teacher does not teach. We get questions and answers to all the homework and tests, but no explanations. I am frustrated! Can anyone explain HOW to come across these answers? Even just one explanation would be greatly appreciated!Question:You are on Homecoming Committee and need to choose a court (3 attendants, 1 queen) from 12 candidates. a. How many ways can you choose a court (ignoring positions)? Answer: 495b. How many ways can you choose a court composed of 3 attendants, 1 queen?Answer: 1980c. How many ways can you choose a court composed of a queen, first runner-up, second runner-up, and a third runner-up?Answer: 11880Question:Two fair dice are tossed and the up face on each die is recorded. Fine the probability of the following event:Event: A 4 appears on at least one of the dice.Answer: 11/36Event: One of the die records an even number.Answer: ??THANK YOU!
 07-25-2007 #2 Mallory Guest   Posts: n/a 18/36 or reduce to get 1/2.
 07-25-2007 #3 Ahmet I Guest   Posts: n/a We did this last year, but I forgot it.I'll take a wild guess, though, and say 27/36 ???
 07-25-2007 #4 polol Guest   Posts: n/a First questiona. No. of ways = 12C4 = 485 waysb. No. of ways = 12C1 x 11C3 = 1980 waysc. No. of ways = 12C1 x 11C1 x 10C1 x 9C1 = 11880 waysSecond questionEvent: A 4 appears on at least one of the dice.P(A 4 appears on at least one of the dice.) = P(4,1) + P(4,2) + P(4,3) + P( 4,4) + P (4,5) + (4,6) + P(1,4) + P(2,4) + ....= (1/6 x 1/6)x11 = 11/36.

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