Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on :

(i) the same day

(ii) consecutive day

(iii) different days.

Total number of possible outcomes = 5 × 5 = 25.

Let the five days of the week be denoted as Monday by M, Tuesday by T, Wednesday by W, Thursday by Th and Friday by F; then :

(i) Favourable cases for employees being absent on same day are : MM, TT, WW, Th Th and FF.

No. of favourable cases = 5.

Required probability = $\dfrac{\text{No. of favourable cases}}{\text{No. of possible cases}} = \dfrac{5}{25} = \dfrac{1}{5}$.

Hence, probability that employees remain absent on same day = $\dfrac{1}{5}$.

(ii) Favourable cases for employees being absent on consecutive day are : MT, TM, TW, WT, W Th, Th W, Th F and F Th.

No. of favourable cases = 8.

Required probability = $\dfrac{\text{No. of favourable cases}}{\text{No. of possible cases}} = \dfrac{8}{25}$.

Hence, probability that employees remain absent on consecutive day = $\dfrac{8}{25}$.

(iii) We know that,

Employees being absent on same day and different day are complimentary events.

∴ P(absent on different days) + P(absent on same day) = 1

⇒ P(absent on different days) = 1 - P(absent on same day)

⇒ P(absent on different days) = 1 - $\dfrac{1}{5} = \dfrac{4}{5}$.

Hence, the probability that employees remain absent on different days = $\dfrac{4}{5}$.