91影视

Given in the question that, Spell-checking software catches 鈥渘onword errors,鈥� which result in a string of letters that is not a word, as when 鈥渢he鈥� is typed as 鈥渢eh.鈥� When undergraduates are asked to write a 250-word essay (without spell-checking), the number X of nonword errors has the following distribution:

We need to find the probability of event 鈥渁t least one nonword error鈥� .

The probability distribution is:

Consider, $X$be the random variable that shows the number of nonword errors. The at least non-word error could be written as $X鈮�1$.

The probability is computed as:

$P(X鈮�1)=1鈭�P(X<1)$

$=1鈭�P(X=0)$

$=1鈭�0.1$

$=0.9$

The probability is$0.9.$

Spell-checking software catches 鈥渘onword errors,鈥� which result in a string of letters that is not a word, as when 鈥渢he鈥� is typed as 鈥渢eh.鈥� When undergraduates are asked to write a 250-word essay (without spell-checking), the number X of nonword errors has the following distribution:

We need to find the probability of events$X鈮�2$and$X<2$

$X鈮�2$means that the non-word errors are $2$or less.

The probabilities could be calculated as:

$P(X鈮�2)=P(X=0)+P(X=1)+P(X=2)$

$=0.1+0.2+0.3$

$=0.6$

Now,

$P(X<2)=P(X=0)+P(X=1)$

$=0.1+0.2$

$=0.3$

Thus, the probabilities are $0.6$ and $0.3$ respectively.