Chapter 13.6, Problem 43E

To determine

To find: the probability of the letters in the word house or in the word phone.

Answer to Problem 43E

The probability that the letters is contained in the word house or in the word phone is $\frac{7}{26}$ .

Explanation of Solution

Given:

A letter is picked at randomfrom the alphabet.

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

Here from the given:

  $\begin{array}{l}P\left(\text{letter is contained in house or phone}\right):\\ \Rightarrow P\left(\text{letter is contained in house}\right)+P\left(\text{letter is contained in phone}\right)-P\left(\text{letter is contained in house and phone}\right).\end{array}$

There are $26$ letters , $5$ in house, $5$ in phone and $3$ common letter between in the two words.

  $P\left(\text{letter is contained in house or phone}\right)=\frac{5}{26}+\frac{5}{26}-\frac{3}{26}=\frac{7}{26}.$

Hence, the probability that the letters is contained in the word house or in the word phone is $\frac{7}{26}$ .