Probability??!!

1) 一枚正常硬幣拋 9 次 , 有幾多次機會會拋出有 6 次公, 3 次字。
( 答案我唔係要個機率 即係0.XXXXX , 而要知一個數字 )
每一種拋出的機會率=(1/2)^9而拋出6次公, 3次字的方法= 9C3種方法機率 =(1/2)^9* 9C3=(1/2)^9*84=0.164 (corr to 3 significant fig.)
2 ) A & B 係2個聽電話人員, 公司得一部電話。EVERYDAY 有40%電話由 A 聽,
有60%電話由 B 聽。NOW, 公司要佢地輪住出trip, so A is out for 20% of his
working time, and B is out for 30% of his working time.

a) find the probability that 3 個連續打黎ge電話係由同一個人聽
P(1) 如果都是由 A 接聽, 而 A 和 B無出 trip=(1-20%)(1-30%)40%^3
=(0.8)(0.7)(0.4)^3
=0.56 * 0.4^3=0.03584 (corr to 4 significant figures)

P(2) 如果都是由A 接聴, 而 B出 trip (那一定由 A接聽)= (1-20%)30%
=0.8*0.3=0.24

P(3) 如果都是由B接聽,而 A 和 B無出 trip=(1-20%)(1-30%)60%^3
=(0.8)(0.7)(0.6)^3
=0.56 * 0.6^3=0.121 (corr to 3 significant figures)

P(4) 如果都是由B 接聴, 而 A出 trip (那一定由 B接聽)= 20%(1-30%)
=0.2*0.7=0.14

3 個連續打黎ge電話係由同一個人聽=P(1)+P(2)+P(3)+P(4)
=0.03584+0.24+0.121+0.14
=0.53684

b) find the probability that 有1 個電話 唔可以answered by the officer (A/B) being called.
P( A & B 都出 trip)
20%*30%=0.2*0.3=0.06