Chapter 13.6, Problem 41E

(a)

To determine

To find: Theprobability of Trina to reach back from where she started the task.

Answer to Problem 41E

The probability to end up where she began is $0.246$ m.

Explanation of Solution

Given:

Trina tossed fair coin.

If its fall heads she walks $10$ meters north.

If its fall tails she walks $10$ meters south.

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

The number of times the coins flipped is:

  $\begin{array}{l}n=\frac{100}{10}=10.\\ P\left(\text{walking north=walking south}\right)=\frac{1}{2}.\end{array}$

To end up the where she began, she walked north and south $5$ times each:

  $\begin{array}{l}P\left(\text{back at her starting point}\right):\\ \Rightarrow C{}_5\times {\left(\frac{1}{2}\right)}^5\times {\left(\frac{1}{2}\right)}^5.\\ \Rightarrow 255\times \frac{1}{32}\times \frac{1}{32}=\mathrm{0.246.}\end{array}$

Hence, the probability to end up where she began is $0.246$ m.

(b)

To determine

To find: the probability within $10$ meters from where she started the journey.

Answer to Problem 41E

The probability within $10$ meter of the starting point is $0.2461$ .

Explanation of Solution

Given:

Trina tossed fair coin.

If its fall heads she walks $10$ meters north.

If its fall tails she walks $10$ meters south.

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

If she is within $10$ meters of the starting point, then the difference between her total distance north and total distance south must lie in the interval $\left(-10,10\right)$ .however ,since she has walked $100$ meters, or $10$ times, she is taken an even number of routes and therefore the only way to she meets the requirement is if she walked $50$ meters north and $50$ meters south. Put of all possible $2^{10}$ outcomes for her walk, that accounts for $C{}_5$ .

The probability within $10$ meter of the starting point is:

  $\begin{array}{l}P\left(\text{within the }10\text{ meter}\right)=\frac{C{}_5}{2^{10}}.\\ \Rightarrow \frac{252}{1024}=\mathrm{0.2461.}\end{array}$

Hence, probability within $10$ meter of the starting point is $0.2461$ .

(c)

To determine

To find: The probability of exactly $20$ meters from the starting point.

Answer to Problem 41E

The exactly $20$ meters from the starting point is $0.41$ .

Explanation of Solution

Given:

Trina tossed fair coin.

If its fall heads she walks $10$ meters north.

If its fall tails she walks $10$ meters south.

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

The number of times the coins flipped is:

  $\begin{array}{l}n=\frac{100}{10}=10.\\ P\left(\text{walking north=walking south}\right)=\frac{1}{2}.\end{array}$

To be exactly $20$ meters from the starting point, she must move $6$ times north and $4$ times south OR $4$ south and $6$ times north.

  $\begin{array}{l}P\left(\text{exactly }20\text{ meters from the staring point}\right):\\ \Rightarrow C{}_6\times {\left(\frac{1}{2}\right)}^6\times {\left(\frac{1}{2}\right)}^4+C{}_4\times {\left(\frac{1}{2}\right)}^4\times {\left(\frac{1}{2}\right)}^6.\\ \Rightarrow 210\times \frac{1}{64}\times \frac{1}{16}+210\times \frac{1}{16}\times \frac{1}{64}.\\ \Rightarrow \mathrm{0.41.}\end{array}$

Hence, exactly $20$ meters from the starting point is $0.41$ .