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

Bob

]]>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

]]>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

]]>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

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