F.5 Maths M1 Distribution(2)

YTC

2015-05-01 9:08 am

1. Two 16-seater mini-buses, A and B, are awaiting at a station. A and B have
already had 11 and 12 seats occupied respectively. Each passenger selects A with probability 4/7 and selects B with probability 3/7. Find the probability that A is filled up first.

My calculation:
http://postimg.org/image/sfy4ye4fn/
http://postimg.org/image/be5awapkj/

Answer: 0.5272

Why the probability is so high? Although selecting A's probability is higher than B, B only need to fill up 4 seats while A need 5 seats. Then the probability of B is fille up first should be higher that A first.

2. A fair dice is thrown a number of times. Find the probability that the first '6'
occurs at the fifth throw and there are three 2's in the first five throws.

My calculation: http://postimg.org/image/gbivherjn/

Answer: 0.0021

What mistake(s) have I made? I have checked many times and changed some
mistakes that I could find. But I still cannot calculate the correct answer orz

Please help. Thank you very much~~

回答 (1)

用戶2053

2015-05-02 2:14 pm

✔ 最佳答案

1)A : 16 - 11 = 5 seats
B : 16 - 12 = 4 seatsP(E) = P(The first 5 passengers select total 5A )
+ P(The first 5 passengers select total 4A1B and the 6th selects A)
+ P(The first 6 passengers select total 4A2B and the 7th selects A)
+ P(The first 7 passengers select total 4A3B and the 8th selects A)= (4/7)⁵
+ 5C4 × (4/7)⁴ (3/7) × 4/7
+ 6C4 × (4/7)⁴ (3/7)² × 4/7
+ 7C4 × (4/7)⁴ (3/7)³ × 4/7= 434176 / 823543 ≈ 0.5272
2)頭四次共擲得三個 2 和一個非2非6 , 第五次擲得 6。
P(E) = 4C3 × (1/6)³ × 4/6 × 1/6 = 16/6⁵ ≈ 0.0021

2015-05-03 00:57:08 補充：
如果你是用符合的組合數目 / 擲五次的全部可能數目來算概率的話,
計算符合的組合數目時會很明暸怎樣去忽略 6(和2) :
即 4C3 × 1³ × (6 - 2) / 6^5 , 1表(2一種忽略1,3,4,5,6) ,
(6 - 2)表(1,3,4,5四種忽略2和6)。

回到原式, 渉及分數時你就醉了: 4C3 × (1/6)³ × 4/6 × 1/6
請看吧 :
(1/6) 表(只擲2一種忽略1,3,4,5,6) , 4/6 表(擲1,3,4,5四種忽略2和6)。

親愛的,你覺悟了嗎? 忽略的都是從分子中減去的,而非從分母中扣除啊!

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