Question

A number is selected from the first 20 natural numbers. Find the probability that it would be divisible by 3 or 7?

(a) \(\frac{19}{46}\)

(b) \(\frac{24}{67}\)

(c) \(\frac{12}{37}\)

(d) \(\frac{7}{20}\)

I got this question during an interview for a job.

My query is from Addition Theorem on Probability in chapter Discrete Probability of Discrete Mathematics

Answer 1

Right option is (d) \(\frac{7}{20}\)

The explanation is: Let X be the event that the number selected would be divisible by 3 and Y be the event that the selected number would be divisible by 7. Then A u B denotes the event that the number would be divisible by 3 or 7. Now, X = {3, 9, 12, 15, 18} and Y = {7, 14} whereas S = {1, 2, 3, …,20}. Since A n B = Null set, so that the two events A and B are mutually exclusive and as such we have,

P(A u B)  =  P(A) + P(B)  ⇒  P(A u B)  =  \(\frac{5}{20} + \frac{2}{20}\)

Therefore,  P(A u B)  =  \(\frac{7}{20}\).