Thread: Coin tossing help View Single Post
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 Quote by MACHO-WIMP okay so the possible outcomes: HH, HT, TH, TT so with exactly one is 2/4. ahhhh that makes sense. so why does my way work?
I'll use the notation P(outcome) to mean "the probability of 'outcome' "

There are two ways to toss exactly one tail. You have have TH or HT. Therefore:

P(exactly one tail) = P(TH OR HT)

Now, these outcomes are mutually exclusive. That means that if one happens, the other one cannot happen, and vice versa. For two events that are mutually exclusive, the probability of one OR the other is just equal to the SUM of the two individual probabilities. In other words:

P(TH OR HT) = P(TH) + P(HT)

I want to emphasize that this rule that you can just sum the probabilities applies only to mutually exclusive events.

Now, the question becomes, what does P(TH) really mean? Well, what it means is:

P(coin 1 is Tails AND coin 2 is Heads)

Similarly, P(HT) = P(coin 1 is Heads AND coin 2 is Tails).

Now, the outcomes of the first coin toss and the second coin toss are independent of each other. What I mean by that is that the outcome of the second coin toss is not influenced by the outcome of the first coin toss in any way. They are not dependent on each other. It turns out that for two independent events, the probability of BOTH of them happening is simply equal to the PRODUCT of the probabilities of the individual events happening. In other words:

P(TH) = P(T)*P(H)

or, written in words:

P(coin 1 is Tails AND coin 2 is Heads) = P(coin 1 is Tails)*P(coin 2 is Heads).

The similar rule applies to P(HT)

Putting all of this together, we get:

P(exactly one tail) = P(TH) + P(HT)

= P(T)*P(H) + P(H)*P(T)

Now, since the coins are both fair, P(T) = P(H) = 1/2. So the expression becomes:

(1/2)*(1/2) + (1/2)*(1/2) = (1/4) + (1/4) = 1/2.