Question

A survey showed that 81​% of adults need correction​ (eyeglasses, contacts,​ surgery, etc.) for their eyesight. If 11 adults are randomly​ selected, find the probability that no more than 1 of them need correction for their eyesight. Is 1 a significantly low number of adults requiring eyesight​ correction?

The probability that no more than 1 of the 11 adults require eyesight correction is

Answer 1

5.579 *
`10^(-7)` or 0.00005579 %

Explanation:

In this question, you select 11 random adults, so the order is not important and we should use combination instead of permutation. There are two different probability here

A= chance that the adults need correction= 0.81

B= chance that the adult doesn't need correction = 0.19

The case that can fulfill the condition of no more than 1 of 11 adults need correction is:

1. 0 adult need correction = 0C11 *
`B^(11)`

2. 1 adult need correction = 1C11 *
`A^(1)` *
`B^(10)`

Then the probability will be:

0C11 *
`B^(11)` +1C11 *
`A^(1)` *
`B^(10)`=

`(11!)/(0!(11-0)!)` *
`0.19^(11)` +
`(11!)/(1!(11-1)!)` *
`0.19^(10)` *0.81 =

5.579 *
`10^(-7)`= 0.00005579 %

The threshold for a significant value that widely used is 5% and the chance is lower than 5%, so no more than 1 adult need correction is a significantly low number of adults requiring eyesight​ correction?