Chapter 5, Problem 19CRP

a.

To determine

Find the probability of the first head to occur in the second trial.

Check whether the probability changes for three tosses.

Find the probability of getting the first head in four tosses.

Mention the probability distribution of the given random variable r.

Answer to Problem 19CRP

The probability of getting a head in the second trial is 0.25.

Yes, the probability change for three tosses is 0.125.

The probability of getting a head in the fourth toss is 0.0625.

The given random variable follows a geometric distribution with $p=\mathrm{0.5.}$

Explanation of Solution

Geometric distribution:

• There should be n independent trials.
• Each trial has two outcomes.
• The probability of success is the same for each trial.
• The random variable n represents the number of trial in which the first success occurs.

The given random variable is based on getting the first head in tossing a coin and it satisfies the above conditions. Hence, it follows a geometric distribution.

The probability mass function of geometric distribution is given as follows:

$P(n)=p{\left(1-p\right)}^{n-1}$

Where, $n$ is the number of binomial trials and $p$ is the probability of success on each trial.

From the given problem, the probability of getting a head while tossing a coin is 0.5. That is, $p=0.5$

$\begin{array}{c}P(n)=(0.5){(1-0.5)}^{n-1}\\ =(0.5){(0.5)}^{n-1}\end{array}$

Therefore, the probability distribution of the given random variable n is $(0.5){(0.5)}^{n-1}$.

The probability of getting a head in the second trial is obtained below:

$\begin{array}{c}P\left(n=2\right)=\left(0.5\right){\left(1-0.5\right)}^{2-1}\\ =\left(0.5\right){\left(0.5\right)}^{2-1}\\ =\left(0.5\right){\left(0.5\right)}^1\\ =0.25\end{array}$

Therefore, the probability of getting a head in the second trial is 0.25.

The probability of getting a head in the third trial is obtained below:

$\begin{array}{c}P\left(n=3\right)=\left(0.5\right){\left(1-0.5\right)}^{3-1}\\ =\left(0.5\right){\left(0.5\right)}^{3-1}\\ =\left(0.5\right){\left(0.5\right)}^2\\ ={\left(0.5\right)}^3\\ =0.125\end{array}$

Therefore, the probability of getting a head in the third trial is 0.125.

The probability of getting a head in the fourth trial is obtained below:

$\begin{array}{c}P\left(n=4\right)=\left(0.5\right){\left(1-0.5\right)}^{4-1}\\ =\left(0.5\right){\left(0.5\right)}^{4-1}\\ =\left(0.5\right){\left(0.5\right)}^3\\ ={\left(0.5\right)}^4\\ =0.0625\end{array}$

Therefore, the probability of getting a head in the fourth trial is 0.0625.

The given random variable follows a geometric distribution with $p=\mathrm{0.5.}$

b.

To determine

Find the probability of getting a head in the fourth trail.

Find the probability of getting a head after four trials.

Answer to Problem 19CRP

The probability of getting a head in the fourth trial is 0.0625.

The probability of getting a head after four trials is 0.0625.

Explanation of Solution

Calculation:

The probability of getting a head in the fourth trial is obtained below:

$\begin{array}{c}P\left(n\right)=\left(0.5\right){\left(0.5\right)}^{n-1}\\ P\left(n=4\right)=\left(0.5\right){\left(1-0.5\right)}^{4-1}\\ =\left(0.5\right){\left(0.5\right)}^{4-1}\\ =\left(0.5\right){\left(0.5\right)}^3\\ ={\left(0.5\right)}^4\\ =0.0625\end{array}$

Therefore, the probability of getting a head in the fourth trial is 0.0625.

The probability of getting a head after four trials is obtained below:

$\begin{array}{c}P\left(n\right)=\left(0.5\right){\left(0.5\right)}^{n-1}\\ P\left(n>4\right)=1-P\left(n\le 3\right)\\ =1-\left[P\left(n=1\right)+P\left(n=2\right)+P\left(n=3\right)+P\left(n=4\right)\right]\\ =1-\left[{\left(0.5\right)}^1+{\left(0.5\right)}^2+{\left(0.5\right)}^3+{\left(0.5\right)}^4\right]\\ =1-0.9375\\ =0.0625\end{array}$

Therefore, the probability of getting a head after four trials is 0.0625.