8 basketball players are to be selected to play in a special game. the players will be selected from a list of 27 players. if the players are selected randomly, what is the probability that the 8 tallest players will be selected?
2006-10-22 09:54:44 · 7 answers · asked by Mr. Rocket Scientist 1 in Science & Mathematics ➔ Mathematics
8!/(27!/19!)=0.000045% or 1 time out of 2220075
2006-10-22 10:06:55 · answer #1 · answered by Anonymous · 3⤊ 1⤋
You have two choices and both lead to the same answer (of course).
1.) Use the Hypgeomdist function in Excel with the arguments (8;8;8;27). The answer you would get is 0,000045%.
2.) The probability of getting one of the tallest efter the first selection is 8/27. At the second selection, you have 7 of the tallest left of a total to be choosen from of 26. I.e. selecting one of the tallest at the first AND one of the remaining tallest at the second attempt is 8/27*7/26 where "*" is times.
Continuing the same reasoning for the remaining 6 tallest the answer is:
8/27*7/26*6/25*5/24*4/23*3/22*2/21*1/20 = 0,000045%
Peter
2006-10-22 17:18:44 · answer #2 · answered by peter n 1 · 2⤊ 0⤋
The first tallest has a chance of 1/27. After he is selected, the second tallest has 1/26 etc. So the chance of the event happening is
1/27 x 1/26 x 1/25..........1/20
= 1/ 8.95^10.
Or more rigorously,
1/ 27!/ 19! = 1/ 8.95^10, as before
2006-10-22 17:21:32 · answer #3 · answered by Anonymous · 1⤊ 0⤋
27 C 8=2220075
Only one of those has the 8 tallest so
1/2220075 = 0.00000045
2006-10-22 17:13:18 · answer #4 · answered by MollyMAM 6 · 2⤊ 0⤋
(8/27)*(7/26)*(6/25)*(5/24)*(4/23)*(3/22)*(2/21)*(1/20)= which works out to a 1/2220075 chance.
2006-10-22 17:11:22 · answer #5 · answered by bruinfan 7 · 3⤊ 0⤋
(27 P 19) / 27! = 0.0000248
2006-10-22 17:06:53 · answer #6 · answered by jacinablackbox 4 · 0⤊ 3⤋
a probability of 38%
2006-10-22 17:06:26 · answer #7 · answered by Anonymous · 0⤊ 3⤋