Complete question :
When Hiroto is writing, there is 0.92 probability that there will be no spelling mistakes on a page. One day, Hiroto writes an essay that is 11 pages long.
Assuming that Hiroto is equally likely to have a spelling mistake on each of the 11 pages, what is the probability that he will have a spelling mistake on at least one of the pages?
Answer:
0.60
Step-by-step explanation:
The question meets the requirement for a binomial probability distribution :
Recall:
P(x = x) = nCx * p^x * q^(n-x)
Given :
Probability of making a spelling mistake = 1 - p(not making) = 1 - 0.92 = 0.08
Hence,
p = 0.08 ; q = 0.92
n = 11
P(x ≥ 1) = 1 - p(x = 0)
P(x = 0) = 11C0 * 0.08^0 * 0.92^11
P(x = 0) = 1 * 1 * 0.3996373778857415671808
P(x = 0) = 0.399
P(x ≥ 1) = 1 - p(x = 0)
P(x ≥ 1) = 1 - 0.399
P(x ≥ 1) = 0.601
P(x ≥ 1) = 0.60
