Question

(a) A weighted coin hasprobability$\frac{\mathbf{2}}{\mathbf{3}}$ ofcoming up heads and probability$\frac{\mathbf{1}}{\mathbf{3}}$of coming up tails. The coin is tossed twice. Let x = number of heads. Set up the sample space for x and the associated probabilities.

(b) Find x and σ.

(c)If in (a) you know that there was at least one tail, what is the probability that both were tails?

Short answer

(a)

$\begin{array}{l}P(x=0)=\frac{1}{4}\\ P(x=1)=\frac{3}{4}\\ P(x=2)=\frac{1}{4}\end{array}$

(b)

$\begin{array}{l}\overline{x}=4/3,\\ \sigma =2/3\end{array}$

(c)

${\text{P}}_{\text{A}}\text{(B)=1/5}$

Step-by-step solution

Given Information.

It has been given that a weighted coin has probability$\frac{2}{3}$ of coming up heads and probability$\frac{1}{3}$ of coming up tails

Definition of Probability. 

Probability means the chances of any event to occur is called it probability.

Find the probability.

a) The sample space S is given as shown below.

$S=\{hh,ht,th,tt\}$

The variable x= "Number of heads" is given by$x=\{0,1,2\}$

The associated probabilities are given as shown below.

$\begin{array}{l}P(x=0)=\frac{1}{4}\\ P(x=1)=\frac{3}{4}\\ P(x=2)=\frac{1}{4}\end{array}$

Find the variance. 

b) The mean value$\overline{x}$is given as shown below.

$\begin{array}{c}\overline{x}=\sum _{i=1}^n{x}_{i}p\left(x_i\right)\\ =0\times \frac{1}{9}+1\times \frac{4}{9}+2\times \frac{4}{9}=\frac{4}{3}\end{array}$

The variance is given as shown below.

$\begin{array}{c}{\sigma }^2=\sum _{i=1}^n{(x_i-\overline{x})}^{2}p\left(x_i\right)\\ ={(0-\frac{4}{3})}^2\left(\frac{1}{9}\right)+{(1-\frac{4}{3})}^2\left(\frac{4}{9}\right)+{(2-\frac{4}{3})}^2\left(\frac{4}{9}\right)\\ =\left(\frac{4}{9}\right)\end{array}$

Find the standard deviation.

$\begin{array}{c}\sigma =\sqrt{\mathrm{var}\left(x\right)}\\ =\sqrt{\left(\frac{4}{9}\right)}\\ =\frac{2}{3}\end{array}$

Step 5:Find the probability. 

c) If we have a condition A= "that at least one of tosses is tail" and we need to find the probability of event B=" both are tails" then conditional probability$P_A\left(B\right)$ is given by the expression mentioned below.

$P_A\left(B\right)\begin{array}{c}{P}_A\left(B\right)=\frac{P\left(AB\right)}{P\left(A\right)}\\ =\frac{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)}{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)+2\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)}\\ =\left(\frac{1}{5}\right)\end{array}$

Hence, the solutions are derived as:

(a)

$\begin{array}{l}P(x=0)=\frac{1}{9}\\ P(x=1)=\frac{4}{9}\\ P(x=2)=\frac{4}{9}\end{array}$

(b)

$\begin{array}{l}\overline{x}=4/3,\\ \sigma =2/3\end{array}$

(c)

${\text{P}}_{\text{A}}\text{(B)=1/5}$