56 employees in an office wear glasses 32 have single vision correction and 24 wear bifocals if 2 employees are selected at random from this group what is the probability that both of them where bifocals what is the probability that both have single vision correction
This is just multiple selection without replacement
P(bi,bi) = 24/56 * 23/55
P(single,single) = 32/56 * 51/55
typo 31/55
To find the probabilities, we need to know the total number of employees wearing glasses and the total number of employees wearing bifocals or single vision correction.
The total number of employees wearing glasses is given as 56.
Let's start with finding the probability that both employees wear bifocals.
We have 24 employees wearing bifocals out of the total 56 employees wearing glasses.
To calculate the probability, we divide the number of outcomes where both employees wear bifocals (24) by the total number of possible outcomes (56 choose 2) when 2 employees are selected from the group.
The formula to calculate the number of combinations is:
nCr = n! / (r!(n-r)!)
Using this formula, we calculate 56 choose 2:
56C2 = 56! / (2!(56-2)!)
= (56 * 55) / (2 * 1)
= 1540
The probability that both employees wear bifocals is:
P(both wearing bifocals) = 24 / 1540
= 0.0156
Now let's find the probability that both employees have single vision correction.
We have 32 employees with single vision correction out of the total 56 employees wearing glasses.
Using the same formula as above, we calculate 56 choose 2:
56C2 = 56! / (2!(56-2)!)
= (56 * 55) / (2 * 1)
= 1540
The probability that both employees have single vision correction is:
P(both having single vision correction) = 32 / 1540
= 0.0208
So, the probability that both selected employees wear bifocals is approximately 0.0156, and the probability that both selected employees have single vision correction is approximately 0.0208.