Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another. What is the probability that both will visit the shop on:(i) the same day (ii) different days (iii) consecutive days

Engineering

Mathematics

Conditional Probability

Question

Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another. What is the probability that both will visit the shop on:
(i) the same day (ii) different days (iii) consecutive days

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Hint Find favourable outcomes and the find probability.

Step 1: Find tatal number of ways.

Total number of days to visit the shop = 6 [Mon to sat]

Total possible outcomes = 6 x 6 = 36

So, two customers can visit the shop in 36 ways.

Step 2: Find favourable outcomes of visiting shop on same day.

E → event of visiting shop on same day.

Number of favourable outcomes = 6 which are (M, M), (T, T), (W, W), (Th, Th), (F, F), (S, S)

Probability P(E)

\frac{Number of favourable outcomes}{Total number of possible outcomes}

P (E) = \frac{6}{36} = \frac{1}{6}

Step 3: Apply property of probability.

E₁ → event of visiting shop on different days but P(E₁) = P(E)

Since, P(E) + P(E) = 1

P(E) = 1 – P(E)

P (E_{1}) = 1 - \frac{1}{6} = \frac{5}{6}

Step 4: Find favourable outcomes of visiting shop on consecutive days.

E₂ → event of visiting shop on consecutive days.

Number of favourable outcomes = 5 which are (M, T) (T, W) (W, Th) (Th, F) (F, S)

P (E) = \frac{5}{36}

Hence, required probabilities are and respectively

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