Question

Set up an appropriate sample space for each of Problems 1.1 to 1.10 and use itto solve the problem. Use either a uniform or non-uniform sample space or try both.

A letter is selected at random from the alphabet. What is the probability that it is one of the letters in the word “probability?” What is the probability that it occurs in the first half of the alphabet? What is the probability that it is a letter after x?

Short answer

The required sample space is $\left(a , b , c , \cdots , z\right)$

The probability that the letter is one of the letters in the word “probability” is $\frac{9}{26}$.

The probability that the letter is occurs in the first half of the alphabet is $\frac{1}{2}$

The probability that the letter is a letter after x is$\frac{1}{13}$.

Step-by-step solution

Definition of Probabilityof alphabets

It is known that there are 26 alphabets, this implies that for selecting a letter there are 26 total possible outcomes So, the probability of a letter is $\frac{\mathbf{1}}{\mathbf{26}}$.

Determination ofthe probability that one of the letters in the word “probability”

There are 26 letters, this implies that for selecting a letter there are 26 total possible outcomes. So, the sample space is expressed as follows,

$\left(a , b , c , \cdots , z\right)$

The probability of each letter is $\frac{1}{26}$.

There is total 9 different letters in the word “probability” namely $\left(\text{p} \text{,} \text{r} \text{,} \text{o} \text{,} \text{b} \text{,} \text{a} \text{,} \text{i} \text{,} \text{l} \text{,} \text{t} \text{,} \text{y}\right)$, this implies that the number of outcomes favourable are 9.

Find the probability that the letter is one of the letters in the word “probability” adding the probabilities of each possible outcomes, that is 9 times $\frac{\mathbf{1}}{\mathbf{26}}$.

$\begin{array}{rcl}p& =& 9\times \frac{1}{26}\\ & =& \frac{9}{26}\\ & & \end{array}$

Thus, the probability that the letter is one of the letters in the word “probability” is$\frac{\mathbf{1}}{\mathbf{26}}$.

Determination ofthe probability that it occurs in the first half of the alphabet

The first half of the alphabet series has 13 letters, this implies that the number of outcomes favourable are 13 and total number of outcomes are 26.

Find the probability that the letter is occurs in the first half of the alphabetby adding the probabilities of each possible outcomes, that is 13 times $\frac{1}{26}$.

$\begin{array}{rcl}p& =& 13\times \frac{1}{26}\\ & =& \frac{1}{2}\\ & & \end{array}$

Thus, the probability that the letter is occurs in the first half of the alphabet is$\frac{1}{2}$

 Step 4: Determination of the probability that it is a letter after x

It can be observed that there are only 2 letters after “x” namely “y” and “z”, this implies that the number of outcomes favourable are 2 and total number of outcomes are 26.

Find the probability that the letter is a letter after x by adding the probabilities of each possible outcomes, that is 2 times $\frac{1}{26}$.

$\begin{array}{rcl}p& =& 2\times \frac{1}{26}\\ & =& \frac{1}{13}\\ & & \end{array}$

Thus, the probability that the letter is a letter after x is$\frac{1}{13}$