Question

(a) One box contains one die and another box contains two dice. You select a box at random and take out and toss whatever is in it (that is, toss both dice if you have picked box 2 ). Let x=number of $\mathbf{3}\mathbf{\text{'}}\mathit{s}$showing. Set up the sample space and associated probabilities for x .

(b) What is the probability of at least one3?

(c) If at least one 3 turns up, what is the probability that you picked the first box?

(d) Find xand.$\mathit{\sigma }$

Short answer

(a)$\begin{array}{c}P\left(0\right)=\left(\frac{55}{72}\right)\\ P\left(1\right)=\left(\frac{2}{9}\right)\\ P\left(2\right)=\left(\frac{1}{72}\right)\end{array}$

(b)The required value is given below

$P(x\ge 1)=\frac{17}{72}$

(c)$P_A\left(B\right)=\left(\frac{6}{17}\right)$

(d)$\begin{array}{c}\overline{x}=0.25\\ \sigma =0.464\end{array}$

Step-by-step solution

Given Information

One box contains one die and another box contains two dice. You select a box at random and take out and toss whatever is in it.

Definition of the Standard deviation.

Standard deviation is the square root of variance.

(a) Set up the sample space.

The value of x is.$x=\{0,1,2\}$

The probabilities are given below.

$\begin{array}{c}P\left(0\right)=\left(\frac{55}{72}\right)\\ P\left(1\right)=\left(\frac{2}{9}\right)\\ P\left(2\right)=\left(\frac{1}{72}\right)\end{array}$

(b) Find the probability for at least one 3 . 

The probability of getting one head is given below.

$\begin{array}{c}P(x\ge 1)=P(x-1)+P(x-2)\\ P(x\ge 1)=1-P(x-0)\\ P(x\ge 1)=1-\frac{55}{72}\\ P(x\ge 1)=\frac{17}{72}\end{array}$

(c) Find the probability that you picked a first box.

Probability is given below.

$\begin{array}{c}{P}_A\left(B\right)=\frac{P\left(AB\right)}{P\left(A\right)}\\ =\frac{0.5\times \left(\frac{1}{6}\right)}{\left(\frac{17}{72}\right)}\\ =\left(\frac{6}{17}\right)\end{array}$

(d) Find the value of mean and deviation.

The value of mean is given below.

$\begin{array}{c}\overline{x}=\sum _{i=1}^n{x}_{i}P\left(x_i\right)\\ =\frac{2}{9}+2\times \frac{1}{72}\\ =0.25\end{array}$

Find the value of deviation is given below.

$\begin{array}{c}{\sigma }^2=\sum _{i=1}^n{(z-z)}^{2}P\left(z_i\right)n\\ ={\left(0-0.25\right)}^2\times \text{t}\frac{55}{72}+{\left(1-0.25\right)}^2\times \frac{2}{9}+{\left(2-0.25\right)}^2\times \frac{1}{72}\\ =\frac{31}{144}\end{array}$

The value of deviation is given below.

$\begin{array}{c}\sigma =\sqrt{\frac{31}{144}}\\ =\frac{\sqrt{31}}{12}\\ =0.464\end{array}$