bi) Assume n nails are needed
P(At least 1 rusty nail) = 1 - P(no rusty nail) .... => 見到at least通常都倒番轉諗
= 1 - (1 - 7.88%) ^ n
1 - (1 - 7.88%) ^ n > 97%
0.9212 ^ n < 0.03 .... => unkonwn係次方 + log
n log(0.9212) < log 0.03
n > 42.7 ... => !!! log 0.9212 < 0 => 要轉符號
Therefore , 43 nails are needed.
bii) No more than 3 nails = 0 nails or 1 nails or 2 nails or 3 nails => Binomial distribution
Probability = (1 - 7.88%) ^ 50 + 7.88% ^ 1 x (1 - 7.88%) ^ 49 x 50C1 + 7.88% ^ 2 x (1 - 7.88%) ^ 48 x 50C2 + 7.88% ^ 3 x (1 - 7.88%) ^ 47 x 50C3
= 0.437605 作者: michaelmadrid 時間: 2018-12-24 12:14 AM