A probability thought problem(統計学の謎): Here's the problem:You are given a bag containing 30 red balls and 60 other balls that are either black or yellow. &nbsp;You don’t know how many black or how many

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Angela

A probability thought problem(統計学の謎):

Here's the problem:

You are given a bag containing 30 red balls and 60 other balls that are either black or yellow. You don’t know how many black or how many yellow balls there are, but that the total number of black balls plus the total number of yellow equals 60. The balls are well mixed so that each individual ball is as likely to be drawn as any other. You must make choices for two successive gambles. First pick A or B and then pick C or D.

(GAMBLE A Receive $100 if you draw a red ball) or (GAMBLE B Receive $100 if you draw a black ball)

(GAMBLE C Receive $100 if you draw a red or yellow ball) or (GAMBLE D Receive $100 if you draw a black or yellow ball)

If you could explain your choice a little that would help me out tremendously since this is a survey question for my homework. Thank you for your time^^

統計学の質問:

袋の中にはビー玉が全部で90声ある。30個は赤で、残りの60は黄色と黒。黄色と黒のビー玉の量が具体的に分からない。仮説例だけど、ギャンバルで一万円もらえる機会があれば、下方で書いている選択肢の中で選んでください

1.勝つ条件は

(A) 赤いビー玉を取り出すこと

(B)黒いビー玉を取り出すこと

1.勝つ条件は

(A) 赤い、または、黄色いビー玉を取り出すこと

(B)黒い、または、黄色いビー玉を取り出すこと

少し理由を説明してくれたら助かります^^

(宿題のためです) 宜しくお願いします(｡•᎑•｡)♡♬

2016年10月27日 05:42

コメント · 4

I would pick A and D.

Reason for picking A: With A you know that there is a guaranteed 30% likelihood of winning so the only gamble you are making is whether you pull out of the of thirty red balls. With option B however you are essentially making two distinct gambles - the first that there are more than 30 black balls (as this would be required for option B to be the better of the two) and that you draw one of these balls. Personally I would prefer to bet on the guaranteed 30% chance. [As a side note, if the individual who handed you the bag is the same one you are making the bets with then they would have motivation to put less than 30 black balls in the bag anyway, making option A an even better choice]

Reason for picking D: Selecting option D means that you have a guaranteed 70% likelihood of winning as we know that all of the balls that are not red are either black of yellow. While there is the potential for this group to be mostly comprised of yellow balls (which would mean C would objectively be the better bet) there is also the potential for there to be mostly black balls. I would again select the more secure option, that being the 70% guaranteed chance of success.

2016年10月27日

I think the answers are in pairs.

If you pick gamble A and win then you should then pick gamble D for your second gamble.

The chance of gamble c winning as a stand-alone bet would be 30 (red balls) + 1-59 (yellow balls, assuming there is at least one yellow ball) or 30 + 0-60 if there are allowed to be zero black or yellow balls whereas the odds for gamble D are 60/90 as a stand-alone bet. These odds seem equal BUT if you removed one red ball when you won on gamble A then that swings the odds in favour of gamble D because the equation is now 29 + 1-59 vs 60/90.

If you pick gamble A and lose, the reverse is true and you should thus pick gamble C.

2016年10月27日

I am not sure if you published your question in order to practice your Japanese or your statistics & probability skills.. However I will try to answer.

The distribution of the chance to get a red ball deals with risk: the probability is known;
while the distributions of the chances to get either a black or a yellow ball deals with uncertainty: the probability is unknown. From symmetric considerations we may assume that we have equal numbers of blacks and yellow, but we don't know that for sure.

Practically I would say that the chances of A and B and of C and D are equal;
however, note that A deals with risk, B with uncertainty, C with a mixture of risk and uncertainty, and D with risk (we know the probability of black or yellow!).

2016年10月27日