Question

In an office. 40% of the staffs offered german. 30% offered spanish and 15% offered both german and spanish. if a staff is selected at random. then find the probability that he has offered german or spanish?

Answer 1

Let the total number of staff be 100.

Staff who offered German = 40% = 40 = n(A)

Staff who offered Spanish = 30% = 30 = n(B)

Staff who offered both German & Spanish = 15% = 15 = n( A intersection B)

The formula to find the union is

n(A U B) = n(A) + n(B) - n(A intersection B)

Substituting the values we get

n(AUB) = 40 + 30 - 15 = 70-15= 55

Hence the probability of a selected staff having offered German or Spanish

= $\frac{55}{100}=55%$ or 0.55

If a staff is selected at random. then the probability that he has offered German or Spanish is 55% or 0.55