If there are nine ladies dancing and the probability of a dancing lady accepting an invitation to dance is .18, then what is the expected number of ladies you would have to ask before one accepts???
2007-12-18 12:33:22 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
For this type of problem, the probability of failure on one attempt is (1 - .18) .82. The probability that after trying n times, you will have failed each time is .82^n. The value of .82^4 is approximately .45, meaning that if you try four times, there is about 55% probability of at least on success.
2007-12-18 12:45:58 · answer #1 · answered by A M Frantz 7 · 0⤊ 0⤋
6
2007-12-18 12:36:18 · answer #2 · answered by Anonymous · 0⤊ 1⤋
Banger dude is correct.
If you were to ask 6, then the expected number of acceptances is 6(0.18) = 1.08.
2007-12-18 12:50:04 · answer #3 · answered by Mr Placid 7 · 0⤊ 0⤋
np= 9 x 0.18=1.62 ladies.(rounding 2)
2007-12-18 12:43:38 · answer #4 · answered by cidyah 7 · 1⤊ 0⤋