Question

A fair coin is flipped $10$times. Find the probability that there is a string of $4$consecutive heads by

(a) using the formula derived in the text;

(b) using the recursive equations derived in the text.

(c) Compare your answer with that given by the Poisson approximation.

Short answer

1. The probability that there is a string of 4 consecutive heads by using the formula derived in the text is $0.2451343$.
2. The probability that there is a string of 4 consecutive heads by using the recursive equations derived in the text is $0.2451172$.
3. By comparing the answer we get$0.2211992$

Step-by-step solution

Given information (part a)

A fair coin is flipped $10$times.

We have to find the probability that there is a string of $4$consecutive heads by using the formula derived in the text.

Explanation (part a)

Define event $E_i$that marks that starting from the $ith$position and on, we have$k$consecutive Heads.

From the inclusion-exclusion formula, we have that the probability that there exist such a sequence is

$P\left({\cup }_{l=1}^7{E}_l\right)=\sum _{r=1}^7(-1{)}^{r+1}\sum _{l_1<\cdots <l_r}P\left(E_{l_1}\dots E_{l_r}\right)$

since we have that $n-k+1=7$. Using the formula from the text (page 151), we have that the probability is equal to

$P\left({\cup }_{l=1}^7{E}_l\right)=\sum _{r=1}^7(-1{)}^{r+1}\left[\left(\begin{array}{c}10-4r\\ r\end{array}\right)+2\left(\begin{array}{c}10-4r\\ r-1\end{array}\right)\right]{\left(\frac{1}{2}\right)}^{5r}$

Use $R$or some similar programming language (code in $R$given below) yields that the answer is$0.2451343.$

Program (part a)

calc=function(r){$c=(-1)^(r+1)*0.5^(5*r)*($choose$(10-4*r,r)+2$* choose$(10-4*r,r-1))$return$\left(c\right)$}

$x=c(1,2,3,4,5,6,7)$

for ($rinx)$

{$c\left[r\right]=$calc$\left(c\right)$}

sum$\left(c\right)$

Final answer (part a) 

The probability that there is a string of consecutive heads by using the formula derived in the text is$0.2451343.$

Given information (part b)

A fair coin is flipped $10$times.

We need to find the probability that there is a string of $4$consecutive heads by using the recursive equations derived in the text.

Explanation (part b)

If we mark with$P_n$the probability that there exist a sequence of $k$consecutive Heads, from the page $151$we have recursive relation

$P_n=\sum _{j=1}^k{P}_{n-j}(1/2{)}^j+(1/2{)}^k$

Where we have that $P_1=P_2=P_3=0$

and $P_4=\frac{1}{16}$

By Writing the code we obtain the answer$0.2451172$

Step 7:Program(part b)

recursion $=$function $\left(n\right)$

$\{$localid="1646907144252" $If(n==0\parallel n==1\parallel n==2\parallel n==3)return\left(0\right)$

localid="1646907151840" $If(n==4)\text{return}\left(0.0625\right)$

localid="1646907163607" $If(n>4)$

localid="1646907158807" $$\begin{gathered} \{Vector=c(1,2,3,4) \\ Summand==c(4) \\ For(jinvector)\{summand\left[j\right]=recursion(n-j)\times 0.5^{\left(j\right)}\} \\ {\text{return (sum(summand)+0.5}}^4\text{}}} \end{gathered}$$

recursion localid="1646907179272" $\left(10\right)$

Final answer (part b)

By using the recursive equations, we get$0.2451172$

Step 9:Given information(part c)

A fair coin is flipped $10$times. We need to compare your answer with that given by the Poisson approximation.

Explanation (part c)

With $L_{10}<4$mark the event that the longest strike of consecutive Heads is shorter than $4$.

From the formula on the page $149$, we have that the Poisson approximation of that probability is

$P(L_{10}<4)\approx$exp$[-\frac{10-4+2}{2^{4+1}}]=e^{-0.25}=0.7788008$

So the probability that exist a strike of length $4$or even longer is

$P(L_{10}\ge 4)-1-P(L_{10}<4)=0.2211992$

Step 11:Final answer(part c)

By the Poisson approximation, we get$0.2211992$