5.3: Tossing a Coin
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- 105835
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Section 1: Tossing a Fair Coin
Next, we will consider the experiment of tossing a fair coin several times and finding the probability of getting a certain number of tails and heads.
A fair coin is a coin with two equally likely outcomes.
Let’s denote the probability of having \(k\) heads among \(n\) tosses as
\(P(kH/n)\)
Let’s find the probability of having no heads among 4 tosses. To do that we are going to list the entire sample space that consists of \(2^4=16\) simple outcomes.
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT,
THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT
The probability of having 0 heads among 4 tosses can be found by dividing the number of outcomes with 0 heads by the size of the sample space.
\(P(0H/4)=\frac{1}{2^4}\)
The probability of having 1 heads among 4 tosses can be found by dividing the number of outcomes with 1 heads by the size of the sample space.
\(P(1H/4)=\frac{4}{2^4}\)
The probability of having 2 heads among 4 tosses can be found by dividing the number of outcomes with 2 heads by the size of the sample space.
\(P(2H/4)=\frac{6}{2^4}\)
The probability of having 3 heads among 4 tosses can be found by dividing the number of outcomes with 3 heads by the size of the sample space.
\(P(3H/4)=\frac{4}{2^4}\)
The probability of having 4 heads among 4 tosses can be found by dividing the number of outcomes with 4 heads by the size of the sample space.
\(P(1H/4)=\frac{4}{2^4}\)
To understand the pattern let’s focus on the probability of having 2 heads among 4 tosses that can be found by dividing the number of outcomes with 2 heads by the size of the sample space.
6 is the number of outcomes with 2H/4 and it is the answer to the question:
How many outcomes with 2H out of 4 tosses?
And it is also the answer to the question:
How many ways to create an outcome with 2H out of 4 tosses?
Which is the same as asking:
How many ways to select 2 tosses out of 4 tosses for H?
So, the 6 is the same as the number of ways to select two objects out of four which we denote as:
\(6=C_2^4\)
It is easy to verify that the numerators in the remaining 4 computations can also be alternatively obtained by using the combinations formula:
\(P(0H/4)=\frac{1}{2^4}=\frac{C_0^4}{2^4}\)
\(P(1H/4)=\frac{4}{2^4}=\frac{C_1^4}{2^4}\)
\(P(3H/4)=\frac{4}{2^4}=\frac{C_3^4}{2^4}\)
\(P(4H/4)=\frac{1}{2^4}=\frac{C_4^4}{2^4}\)
To summarize, the probability of having \(k\) heads among \(4\) tosses can be found by using the formula:
\(P(kH/4)=\frac{C_k^4}{2^4}\)
And in general, the probability of having \(k\) heads among \(n\) tosses can be found by using the formula:
\(P(kH/n)=\frac{C_k^n}{2^n}\)
Find the probability of 5 heads among 9 tosses of a fair coin.
Solution
According to the formula it is
\(P(kH/n)=\frac{C_k^n}{2^n}=\frac{C_5^9}{2^9}=\frac{126}{512}=24.6\%\)
We discussed the experiment of tossing a fair coin several times and finding the probability of getting a certain number of tails and heads. The significance of this result is that many experiments in real life can be modeled by a toss of a fair coin.
Section 2: Tossing an Unfair Coin
An unfair coin is a coin with two not equally likely outcomes.
Let’s discuss the differences between a fair coin and an unfair coin. The outcomes of a fair coin are equally likely and since there are only two of them, probability of each one is 0.5. The outcomes of an unfair coin are not equally likely and hence the probability of each outcome is unknown unless some information is given.
Let’s find the \(P(T)\) and \(P(H)\) if \(T\) is three times more likely than \(H\).
Solution
The given information can be interpreted as the odds in favor of \(T\) are 3:1 and therefore the probability of \(T\) is 0.75 and the probability of \(H\) is 0.25, i.e.:
\(P(T)=\frac{3}{3+1}=0.75\) and \(P(H)=\frac{1}{3+1}=0.25\)
Let’s find the \(P(T)\) and \(P(H)\) if \(T\) is four times less likely than \(H\).
Solution
The given information can be interpreted as the odds in favor of \(T\) are 1:4 and therefore the probability of \(T\) is 0.20 and the probability of \(H\) is 0.80, i.e.:
\(P(T)=\frac{1}{1+4}=0.20\) and \(P(H)=\frac{4}{1+4}=0.80\)
Let’s consider the sample space of tossing a coin four times:
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT,
THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT
Let’s denote \(P(H)=p\) and \(P(T)=q\). Since the outcomes are not equally likely, let’s find the probabilities of each of the simple outcomes. Since the outcomes of different tosses are independent the probability of 4 Heads can be found using the special multiplication rule and is equal to \(p^4\).
\(P(HHHH)=P(H)\cdot P(H)\cdot P(H)\cdot P(H)=p\cdot p\cdot p\cdot p=p^4\)
Similarly, we can find the probabilities of all simple outcomes.
\(P(HHHT)=P(H)\cdot P(H)\cdot P(H)\cdot P(T)=p\cdot p\cdot p\cdot q=p^3 q^1\)
\(P(HHTH)=P(H)\cdot P(H)\cdot P(T)\cdot P(H)=p\cdot p\cdot q\cdot p=p^3 q^1\)
\(P(HHTT)=P(H)\cdot P(H)\cdot P(T)\cdot P(T)=p\cdot p\cdot q\cdot q=p^2 q^2\)
\(\vdots\)
\(P(TTTH)=P(T)\cdot P(T)\cdot P(T)\cdot P(H)=q\cdot q\cdot q\cdot p=p^1 q^3\)
\(P(TTTT)=P(H)\cdot P(H)\cdot P(H)\cdot P(H)=p\cdot p\cdot p\cdot p=p^4\)
Note that the probability of a simple outcome depends only on the number of heads and tails in the simple outcome!
Let’s denote the probability of having \(k\) heads among \(n\) tosses in the following way
\(P(kH/n)\)
Let’s find the following probabilities. The probability of having 0 heads among 4 tosses can be found by finding the probability of all tails which is \(1q^4\):
\(P(0H/4)=P(TTTT)=1q^4\)
The probability of having 1 heads among 4 tosses can be found by the special addition rule and is equal to \(4pq^3\):
\(P(1H/4)=P(HTTT)+P(THTT)+P(TTHT)+P(TTTH)=4pq^3\)
Similarly, the probability of having 3 heads among 4 tosses can be found by the special addition rule and is equal to \(4p^3q\):
\(P(3H/4)=P(THHH)+P(HTHH)+P(HHTH)+P(HHHT)=4p^3q\)
Finally, the probability of having 4 heads among 4 tosses can be found by finding the probability of all heads which is \(p^4\).
\(P(4H/4)=P(HHHH)=1p^4\)
It is not hard to recognize the coefficients of the expressions as we have already seen these numbers before! So we can find the probability of having 2 heads among 4 tosses without writing out all simple outcomes in the following way:
\(P(2H/4)=P(HHTT)+\cdots+P(TTHH)=6p^2q^2\)
In summary, if \(P(H)=p\) and \(P(T)=q\) then the probability of having \(k\) heads among \(n\) tosses is given by the following formula:
\(P(kH/n)=C_k^n p^k q^{n-k}\)
Find the probability of having 6 heads among 10 tosses if the heads are 7 times likelier than tails.
Solution
First, find \(p=P(H)\) and \(q=P(T)\) using the fact that the odds in favor of \(H\)are 7 to 1:
\(P(H)=\frac{7}{7+1}=\frac{7}{8}\) and \(P(T)=\frac{1}{7+1}=\frac{1}{8}\)
Now, we can find the probability of having 6 heads among 10 tosses by using the formula:
\(P(6H/10)=C_6^{10}(\frac{7}{8})^6 (\frac{1}{8})^{10-6}=210\cdot0.4488\cdot0.0002=2.3\%\)
In conclusion, if we let the probability of heads and tails to be equal to ½ then the formula for an unfair coin becomes the formula for a fair coin:
\(P(kH/n)=C_k^n (\frac{1}{2})^k (\frac{1}{2})^{n-k}=\frac{C_k^n}{2^n}\)
We discussed the experiment of tossing an unfair coin several times and finding the probability of getting a certain number of tails and heads. The significance of this result is that many experiments in real life can be modeled by a toss of an unfair coin.