Question

On a page of a telephone directory, there are 200 telephone numbers. The frequency distribution of the digits at their units place is given below:

Unit digit	0	1	2	3	4	5	6	7	8	9
Frequency	22	26	22	22	20	10	14	28	16	20

Without looking at the page, a number is chosen at random from the page. What is the probability that the digit at the unit's place of the number chosen is greater than 6?

Answer 1

We have, the frequency distribution of the digits at their units place:

Unit digit	0	1	2	3	4	5	6	7	8	9
Frequency	22	26	22	22	20	10	14	28	16	20

As, the frequency of getting unit digit geater than (i.e. 7, 8 and 9) = 28 + 16 + 20 = 64

$$\begin{gathered} \mathrm{So},\mathrm{the}\mathrm{required}\mathrm{probability}=\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}} \\ =\frac{64}{200} \\ =\frac{8}{25} \end{gathered}$$

Hence, the probability that the digit at the unit's place of the number chosen is greater than 6 is $\frac{8}{25}$.