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Coin tossing help

by MACHO-WIMP
Tags: coin, tossing
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MACHO-WIMP
#1
Mar13-12, 09:26 PM
P: 43
1. The problem statement, all variables and given/known data
two pennies are tossed: find the probabilities for tossing exactly one tail.


2. Relevant equations
n/a


3. The attempt at a solution
I got 1/2 and the book agrees with me, but it doesn't show any work. and i need to know if i'm doing it the correct way. so can someone please help me with how to do it?
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cepheid
#2
Mar13-12, 09:32 PM
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Quote Quote by MACHO-WIMP View Post
1. The problem statement, all variables and given/known data
two pennies are tossed: find the probabilities for tossing exactly one tail.


2. Relevant equations
n/a


3. The attempt at a solution
I got 1/2 and the book agrees with me, but it doesn't show any work. and i need to know if i'm doing it the correct way. so can someone please help me with how to do it?
Why don't you show us what you did, and we can tell you if you are doing it right?

Just write down all possible outcomes of the two coin tosses, and count how many of them are outcomes in which exactly one coin shows tails. The probability of tossing exactly one tails is just the number of outcomes with one tails divided by the total number of all possible outcomes.

The above assumes that all outcomes are equally likely, which is true if both coins are fair.
MACHO-WIMP
#3
Mar13-12, 09:39 PM
P: 43
well i did:

(1/2)^2+(1/2)^2=1/2

I'm not sure why I did this, but I remembered doing it in the previous section.

cepheid
#4
Mar13-12, 09:54 PM
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Coin tossing help

Quote Quote by MACHO-WIMP View Post
well i did:

(1/2)^2+(1/2)^2=1/2

I'm not sure why I did this, but I remembered doing it in the previous section.
Did you try it my way?

Do that first, and then I'll explain why your way works.
MACHO-WIMP
#5
Mar13-12, 09:58 PM
P: 43
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?
cepheid
#6
Mar13-12, 10:12 PM
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Quote Quote by MACHO-WIMP View Post
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.
MACHO-WIMP
#7
Mar13-12, 10:25 PM
P: 43
Thank you so much cepheid!


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