Q.2.13 - A First Course in Probability

Question

Suppose that a person chooses a letter at random from$R E S E R V E$ and then chooses one at random from $V E R T I C A L$. What is the probability that the same letter is chosen? 

Step-by-Step Solution

Answer

The required probability is $P\left(\text{ the same letter is chosen }\right)=\frac{3}{28}$.

Step 1: Given Information

A letter at random from $R E S E R V E$ and then chooses one at random from $V E R T I C A L$.

Step 2: Explanation

 Experiment: Choose a letter form $\left\{R_1,E_1,S_1,E_2,R_2,V_1,E_3\right\}$ and then one from the $\left.\left\{V_1,E_1,R_1,T_1,I_1,C_1,A_1,L_1\right\}\right\}$

Find: the probability that the same letter is chosen.

Outcome space $S$ contains every pair of a letter from the first word and a letter from the second word.

If all events in $S$ are considered equally likely, probability of event $A\subseteq S$ is:

$P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}$

where $\left|X\right|$ denotes the number of elements in $X$.

Step 3: Explanation

The number of elements in $S$ :

$7$ choices for the first letter, and $8$ choices for the second letter. - $\left|S\right|=56$

The event $A$ - the same letter is chosen from both words.

A is an union of three mutually exclusive events $A=A_R\cup A_E\cup A_V·A_R,A_E,A_V$ are events where two Rs, two Es or two Vs are chosen, respectively.

$$\begin{gathered} A_R=\left\{R_1{R}_1,R_2{R}_1\right\}\subseteq S,\text{ and }\left|A_R\right|=2\Rightarrow P\left(A_R\right)=\frac{2}{56}=\frac{1}{28} \\ {A}_E=\left\{E_1{E}_1,E_2{E}_1,E_3{E}_1\right\}\subseteq S,\text{ and }\left|A_E\right|=3\Rightarrow P\left(A_E\right)=\frac{3}{56} \\ {A}_V=\left\{V_1{V}_1\right\}\subseteq S,\text{ and }\left|A_V\right|=1\Rightarrow P\left(A_V\right)=\frac{1}{56} \end{gathered}$$

Since $A_R,A_E,A_V$ are mutually exclusive

$P\left(A\right)=P\left(A_R\right)+P\left(A_E\right)+P\left(A_V\right)=\frac{6}{56}=\frac{3}{28}$