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I have an interesting question that I certainly don't know how to solve it. I've already read many topics on probability, eg: Probability that someone will pick a red ball first? and Comparing probabilities of drawing balls of certain color, with and without replacement etc. But unfortunately, I can't apply the same methodology in this case and get the right answer from the given ones (it seems I'm really silly one). So here is the question:

There are 5 balls in a bucket: green, blue, red, orange and black. Each turn you take a random ball from the bucket. What is a probability that at 2nd turn you will pick blue ball? The answers:

  • 1/2
  • 2/3
  • 1/3
  • 2/5
  • 1/5

The first way I thought is to add probability of each turn like this: $\frac{1}{5} + \frac{1}{4}$ - 1/4 because at 2nd turn we have only four balls. However, the answer become $\frac{9}{20}$ which is not correct.

I know there is something to do with either factorial or combination (just my assumption).

Community
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Tilqur
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  • There is no such an information, but I think it doesn't. But what if doesn't or do? – Timur Fayzrakhmanov Apr 20 '16 at 13:16
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    You want probability of choosing on second turn so first turn you want probability of not choosing green so 4/5 then times the probability of choosing green on second turn 1/4 so we get 1/5 as our answer – Michael N Apr 20 '16 at 13:21
  • The quantity 1/5 + 1/4 that you computed is the probability that on one of the first two turns you pick the green ball. – Gareth McCaughan Apr 20 '16 at 16:59

2 Answers2

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you need the blue ball to be in the bucket after first pick : probability $\frac{4}{5}$

you need to pick the blue ball among the 4 remaining balls : probability $\frac{1}{4}$

Global probability : $$\frac{1}{4}\times \frac{4}{5} = \frac{1}{5}$$

You can also see it this way : The 5 balls are exactly the same so the probability to choose any ball at 2nd turn is $\frac{1}{5}$ for each ball.

leonbloy
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Fabich
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  • Hm..looks nice! Could you explain why sometimes probabilities are added and sometimes multiplied. What is the reason? – Timur Fayzrakhmanov Apr 20 '16 at 13:20
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    @TimurFayzrakhmanov One basic rule can be AND = multiply, OR = add. Here we want to pick a non-blue ball at first AND choose a blue ball at second so we multiply. If we want to choose blue ball OR red ball then we add the probabilities and we find 2/5. (this is a little naive and will not work for complex situations) – Fabich Apr 20 '16 at 13:29
  • I had just written up a more long-winded version of the second part of this answer. Wish I'd noticed that earlier. Well done. – David K Apr 20 '16 at 13:40
  • I don't know how to accept the answer..When I posted it I posted it using just nickname and email. – Timur Fayzrakhmanov Apr 20 '16 at 14:08
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    @TimurFayzrakhmanov: If you are Tilqur, then you should use the contact form and select “I need to merge user profiles” to have your accounts merged.  In order to merge them, you will need to provide links to the two accounts.  For your information, these are http://math.stackexchange.com/users/333152/tilqur and http://math.stackexchange.com/users/333154/timur-fayzrakhmanov.  You’ll then be able to edit and accept answers to this question. – Scott - Слава Україні Apr 20 '16 at 18:05
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If you choose with replacement:

$$\frac15$$

If you choose without replacement:

$$\left(1-\frac15\right)\cdot\frac14$$

barak manos
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