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## Topic review (newest first)

Agnishom
2013-07-03 12:03:53

THanks

bob bundy
2013-07-03 05:52:10

hi

I get that too by my own method and a general formula too.

Bob

bobbym
2013-07-03 05:15:27

Agnishom
2013-07-03 03:42:11

Then?

bobbym
2013-07-03 03:37:10

You can not add them up.

Agnishom
2013-07-03 03:35:49

#### Agnishom wrote:

The Probability that the 1st Person chose dish X is 1/5.
The Probability that the 2nd Person chose dish X is 1/5.
The Probability that the 3rd Person chose dish X is 1/5.
Therefore, the probability that anyone has chosen X before is 3/5.

That method is faulty.  If there were 5 people before D, you would end up with 5/5 and no chance for D.

I'll work on an analytic method.

Bob

Ya,  so... What to do?

Hi bobbym,

They ask me for a + b too

bobbym
2013-07-03 03:33:21

Must be different for me and you then. Mine requests a + b in a / b.

bob bundy
2013-07-03 03:32:26

#### Agnishom wrote:

The Probability that the 1st Person chose dish X is 1/5.
The Probability that the 2nd Person chose dish X is 1/5.
The Probability that the 3rd Person chose dish X is 1/5.
Therefore, the probability that anyone has chosen X before is 3/5.

That method is faulty.  If there were 5 people before D, you would end up with 5/5 and no chance for D.

I'll work on an analytic method.

Bob

Agnishom
2013-07-03 03:29:33

I would say yes, because the answer you gave me works

bobbym
2013-07-03 03:22:01

Is that the exact question?

Agnishom
2013-07-03 03:20:55

Consider that the 4th person chose dish X.

The Probability that the 1st Person chose dish X is 1/5.
The Probability that the 2nd Person chose dish X is 1/5.
The Probability that the 3rd Person chose dish X is 1/5.
Therefore, the probability that anyone has chosen X before is 3/5.

So, the possibility that the dish is previously not chosen is 2/5

bobbym
2013-07-03 03:16:58

Although I have the exact answer by experimental methods I am still working on a good solid method that would please an olympic dude.

Agnishom
2013-07-03 03:10:27

New Problem There are 5 kinds of dishes i, ii, iii, iv, and v. Four Customers A, B, C and D choose a dish at random. What is the probability that person D chooses a previously unordered dish?

bobbym
2013-07-03 02:59:37

Yes, if you are signed in as a member.

Agnishom
2013-07-03 02:47:14

The Featured Solution: https://brilliant.org/i/b6ozMJ/