A question about probability

👨🏻‍🏫 學術台數學

pollyconstant

2012-02-05 11:28 am

Suppose that an ordinary deck of 52cards is shuffled and the cards are then turned over one at a timeuntil the first ace appears. Given that the first ace is the 20thcard to appear, what is the conditional probability that the cardfollowing it is the ace of spades? Ans: 3 / 128

更新1:

I don't understand the first incidence has the probability of 3/4 because you are choose three ace out of 33 cards.

更新2:

I still do not understand. Why 每張牌在任何次序上抽中的機率都是一樣的﹐即1/52。例如紅心ace在第1張﹐第2張﹐...第52張抽中的機率都是1/52. They are not putting the card back to the deck.

更新3:

I have asked my teacher and the solution should be... A = {the first 19 cards are not ace} B = {the 20th card is an ace} C = {the 21st card is an ace of spade} D = {the 20th card is an ace other than ace of spade} P(A) = 48/52 * 47/51...., P(B) = 4/33 P(C) = 1/32, P((A and C and D) | (A and B)) wil

更新4:

P(C) = 1/ 32 P(D) = 3/33 P((A and C and D) | (A and B)) = P(A) x P(C) x P(D) / (P(A) x P(B)) = 3/128

回答 (2)

myisland8132

2012-02-05 8:52 pm

✔ 最佳答案

最簡單明暸是用定義去做。

令事件 A: 第20張卡是ace。B：第21張卡是黑桃ace

P(B | A)

= P(AB)/P(A)

= (3/52)(1/32) /(4/52)

= 3/128

注意：

1 每張牌在任何次序上抽中的機率都是一樣的﹐即1/52。例如紅心ace在第1張﹐第2張﹐...第52張抽中的機率都是1/52

2 事件A和B同時發生當且僅當第20張抽中的是階磚﹐梅花或紅心ace﹐而第21張抽中的是黑桃ace

👍 0 👎 0

用戶25975

2012-02-05 5:36 pm

For the given situation to happen, then first ace appeared should not be the spade one, so this incident has a probability of 3/4.
Then after the 20th card (first ace) turned over, there still remains 32 cards unturned.

Hence the probability that the next one is ace of spades is 1/32.

Thus, the required probability is 3/4 x 1/32 = 3/128

2012-02-05 20:15:49 補充：
Becos the given condition is that the first ace appears at the 20th card, so there are 4 possibilities for this ace card and 3 of them are NOT spade

參考: 原創答案

👍 0 👎 0

收錄日期: 2021-04-26 19:17:14
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120205000051KK00163

檢視 Wayback Machine 備份