F.5 Maths Probability

YTC

2015-06-30 3:18 am

In a game, there are 2 red balls and 18 white balls in a bag. A player will draw 2
balls from the bag randomly. If at least 1 red ball is drawn, the player wins;
otherwise, the player needs to put back the balls drawn and draws 2 balls
randomly from the bag again until at least 1 red ball is drawn.
Find the probability that a player will win in a particular draw.

I think that "the player will win in a particular draw" means "the player will win the game", so the prob. should be 1.
However, the required prob. should be 1-P(No red balls are drawn).

I want to ask what is the meaning of "a particular draw" ??
What is my misunderstanding?

Thank you!!

回答 (1)

50418129

2015-06-30 7:07 pm

✔ 最佳答案

P(No red balls are drawn)
= P(1st round no red) x P(2nd round no red) x P(3rd round no red) x ...
= (18C2 / 20C2) x (18C2 / 20C2) x (18C2 / 20C2) x ...
= lim (n->inf) (18C2 / 20C2)^n
= 0

thus, the required prob. = 1 - P(No red balls are drawn) = 1 - 0 = 0
no misunderstanding.

2015-06-30 11:22:56 補充：
Sorry, I think a particular dram means "指定的一次"
So, the required prob. = 1 - P(No red balls are drawn) = 1 - 18C2 / 20C2

2015-07-02 11:33:52 補充：
可以是指定第二次
P(第二次win) 也是 1 - 18C2 / 20C2

P(第一次lose及第二次win) 才是 (18C2 / 20C2) x (1 - 18C2 / 20C2)

2015-07-02 11:46:22 補充：
P(某指定一次lose) = P(兩個白球) = 18C2 / 20C2 = 0.805
P(某指定一次win) = P(最少一次紅球) = 1 - 18C2 / 20C2 = 0.195
P(第二次才win) = 0.805 x 0.195
P(第三次才win) = 0.805 x 0.805 x 0.195
P(第四次才win) = 0.805 x 0.805 x 0.805 x 0.195

參考: knowledge

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