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Simple coins (Posted on 2002-04-09) Difficulty: 2 of 5
I toss two coins and look at the outcome. I then tell you that at least one of the coins is showing up as "tails". What is the chance that the other one is showing "tails" as well?

(from techInterview.org)

See The Solution Submitted by art    
Rating: 3.8750 (16 votes)

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re(4): Solution Using Real Prob/Stats | Comment 27 of 45 |
(In reply to re(3): Solution Using Real Prob/Stats by SilverKnight)

BTW, Eberhard, your math is correct:

Eberhard wrote:
On Independence-

In this particular example the probability of a particular outcome (which we will call flip one or F1), that is to say H or T, does not change the probability of the outcome of F2 (flip 2).
i.e. when P(F1)>0, P(F1|F2) = P(F2)
indicating independence
Furthermore- the multiplication rule here become

P(F1 intersect F2)=P(F1)P(F2|F1)=P(F1)P(F2)
Which implies that when P(F2)>0, then

P(F1|F2)= (P(F1 intersect F2)/P(F2)) = (P(F1)P(F2))/P(F2)= P(F1)

_______________________

What is not correct is your understanding of the problem.

The problem did not say, "I tell you that the FIRST coin flipped is a tails, what is the probability that the OTHER coin flipped is tails" (in which case your mathematics would apply and the answer would be 1/2).

The problem can be restated as "I tell you that one of the following is true: either the FIRST coin flipped is tails and the SECOND is heads -- or -- the SECOND coin flipped is tails and the FIRST is heads -- or -- both coins are tails. What is the chance that both coins are tails?"

Given this equivalent restatement, the mathematics is different and leads to 1/3.
  Posted by SilverKnight on 2003-12-03 16:10:30

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