Question

Spell-checking Spell-checking software catches “nonword errors,” which result in a string of letters that is not a word, as when “the” is typed as “teh.” When undergraduates are asked to write a 250-word essay (without spell-checking), the number Y of nonword errors in a randomly selected essay has the following probability distribution

Part (a). Write the event “one nonword error” in terms of Y. Then find its probability.

Part (b). What’s the probability that a randomly selected essay has at least two nonword errors?

Short answer

Part (a) 20%

Part (b) 70%

Step-by-step solution

Part (a) Step 1. Given information.

Value $y_i$	0	1	2	3	4
Probability $p_i$	0.1	??	0.3	0.3	0.1

Part (a) Step 2. The event “one nonword error” in terms of Y. 

Let us refer to the missing probability as x.

Value $x_i$	0	1	2	3	4
Probability $p_i$	0.1	x	0.3	0.3	0.1

Y represents the number of nonword errors in a randomly chosen essay, implying that "one nonword error" is represented by "Y = 1."

If all probabilities are between 0 and 1, the probability distribution is valid (including).

The total of the probabilities of each outcome equals one.

This implies that the sum of all probabilities in the preceding table must equal 1.

$$\begin{gathered} 0.1+x+0.3+0.3+0.1=1 \\ 0.8+x=1 \end{gathered}$$

Take 0.8 off each side

role="math" localid="1653985297341" $$\begin{gathered} x=1-0.8 \\ x=0.2 \end{gathered}$$

We can see in the table above that x represents the probability that Y is equal to 1.

$$\begin{gathered} P\left(Y=1\right)=0.2 \\ P\left(Y=1\right)=20\% \end{gathered}$$

Part (b) Step 2. The probability that a randomly selected essay has at least two nonword errors. 

If two events cannot occur at the same time, they are disjoint or mutually exclusive.

Addition rule for events that are disjoint or mutually exclusive:

$P\left(AUB\right)=P\left(A\right)+P\left(B\right)$

Value $x_1$	0	1	2	3	4
Probability $p_i$	0.1	x	0.3	0.3	0.1

In the above table, we can calculate the probability of 0, 2, 3, and 4 nonword errors are:

role="math" localid="1653985957678" $$\begin{gathered} P(Y=2)=0.3 \\ P(Y=3)=0.03 \\ P(Y=4)=0.1 \end{gathered}$$

The two events are mutually exclusive because it is not possible to obtain two different numbers of nonword errors for the same piece of text.

For mutually exclusive events, apply the addition rule.

$$\begin{gathered} P(Y\ge 2)=P(Y=2)+P(Y=3)+P(Y=4) \\ P(Y\ge 2)=0.3+0.3+0.1 \\ P(Y\ge 2)=0.7 \\ P(Y\ge 2)=70\% \end{gathered}$$