Chapter 6.7, Problem 15E

a.

To determine

To find: the number of e’s and a’s on the page containing 300 letters.

Answer to Problem 15E

Number of e’s are on the page is 40 and Number of a’s are on the page is 25.

Explanation of Solution

Given information :

The tables give the probability that a letter chosen at random from a page of English text, is given below:

Letter	Probability
e	0.131
t	0.104
a	0.081
s	0.061

Formula used:

The probability of event A is obtained as:

  $P\left(A\right)=\frac{\text{Number of favourable outcome}}{\text{Number of possible outcome}}$

Calculation :

Let $P\left(e\right)\text{ and }P\left(a\right)$ be the probability of getting e’s and a’s if a letter is chosen at random.

From the table,

  $P\left(e\right)=0.131\text{ and }P\left(a\right)=0.081$

Number of possible outcomes is 300 letters.

Therefore,

  $\begin{array}{l}P\left(e\right)=\frac{\text{Number of }e\text{'}\text{s are on the page}}{\text{Number of letters on the page}}\\ 0.131=\frac{\text{Number of }e\text{'}\text{s are on the page}}{\text{300}}\\ \text{Number of }e\text{'}\text{s are on the page}=0.131\times 300\\ \text{Number of }e\text{'}\text{s are on the page}=39.3\\ \text{Number of }e\text{'}\text{s are on the page}\approx 40\end{array}$

And,

  $\begin{array}{l}P\left(a\right)=\frac{\text{Number of }a\text{'}\text{s are on the page}}{\text{Number of letters on the page}}\\ 0.081=\frac{\text{Number of }a\text{'}\text{s are on the page}}{\text{300}}\\ \text{Number of }a\text{'}\text{s are on the page}=0.081\times 300\\ \text{Number of }a\text{'}\text{s are on the page}=24.3\\ \text{Number of }a\text{'}\text{s are on the page}\approx 25\end{array}$

Hence,

Number of e’s are on the page is 40 and Number of a’s are on the page is 25.

b.

To determine

To find: the experimental probability that a letter randomly chosen from the sentence is a t and the probability that it is a s .

Answer to Problem 15E

Explanation of Solution

Given information :

The tables give the probability that a letter chosen at random from a page of English text, is given below:

Letter	Probability
e	0.131
t	0.104
a	0.081
s	0.061

The two sentence contain 60 letters.

Formula used:

The probability of event A is obtained as:

  $P\left(A\right)=\frac{\text{Number of favourable outcome}}{\text{Number of possible outcome}}$

Calculation :

Let $P\left(t\right)\text{ and }P\left(s\right)$ be the probability of getting t’s and s’s if a letter is chosen at random.

Number of possible outcomes is 60 letters.

Total number of letter t’s is 5.

  $\begin{array}{l}P\left(t\right)=\frac{\text{Number of }t\text{'}\text{s are on the page}}{\text{Number of letters on the page}}\\ P\left(t\right)=\frac{5}{\text{60}}\\ P\left(t\right)=0.083\end{array}$

Total number of letter s’s is 8.

  $\begin{array}{l}P\left(s\right)=\frac{\text{Number of }s\text{'}\text{s are on the page}}{\text{Number of letters on the page}}\\ P\left(s\right)=\frac{8}{\text{60}}\\ P\left(s\right)=0.133\end{array}$

From the table,

  $P\left(t\right)=0.104>0.083\text{ and }P\left(s\right)=0.061<0.133$

Hence,

The experimental probability of choosing t’s is lesser than the probability of choosing t’s

provided in table and experimental probability of choosing s’s is greater than the probability

of choosing t’s provided in table