Question

Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each day is equally likely to visit the shop on another day. What is the probability that both will visit the shop on:
(i) The same day
(ii) Consecutive days
(iii) Different days

Answer 1

Hint:We assume two variables denoting each customer and find the total possibilities of visiting a particular shop. We write the possibilities in the form of ordered pairs where we write the first element as a possibility for one customer and second element as a possibility for the second customer. Using the formula for probability which is the number of favorable outcomes divided by the total number of outcomes we find probability in each of three cases.


Complete step-by-step answer:
We have two customers Shyam and Ekta. Let us denote Shyam by S and Ekta by E. Both customers visit the shop between the days Tuesday, Wednesday, Thursday, Friday and Saturday.
Total number of days of the week in which customers visit the shop is 5.
Then we draw a table for possibilities of them visiting the particular shop on days of the week from Tuesday to Saturday.
We denote the possibilities as a set of ordered pairs \[(x,y)\] where x denotes the day shyam visited the shop and y denotes the day Ekta visited the shop.
Here we denote the days by:
Tuesday -T, Wednesday -W, Thursday -TH, Friday -F and Saturday -S

(Shyam) (Ekta)	Tuesday	Wednesday	Thursday	Friday	Saturday
Tuesday	(T,T)	(W,T)	(TH,T)	(F,T)	(S,T)
Wednesday	(T,W)	(W,W)	(TH,W)	(F,W)	(S,W)
Thursday	(T,TH)	(W,TH)	(TH,TH)	(F,TH)	(S,TH)
Friday	(T,F)	(W,F)	(TH,F)	(F,F)	(S,F)
Saturday	(T,S)	(W,S)	(TH,S)	(F,S)	(S,S)

We see total possible outcomes are 25.
Now we solve each part separately.

(i) The same day:
In this case we take all possibilities where both Shyam and Ekta visit the shop on same day.
So, we take a set of those ordered pairs from the table where both first and second elements are equal.
Favorable outcomes are: (T,T), (W,W), (TH,TH), (F,F), (S,S)
So the number of favorable outcomes is 5.
We use the formula for probability of an event which is given by the number of favorable outcomes divided by the total number of outcomes.
Probability that Shyam and ekta visit shop on same day \[ = \dfrac{5}{{25}}\]
Cancel out common factors from numerator and denominator.
Probability that Shyam and ekta visit shop on same day \[ = \dfrac{1}{5}\]

(ii) Consecutive days:
In this case we take all possibilities where both Shyam and Ekta visit the shop on consecutive days.
So, we take a set of those ordered pairs from the table where both first and second elements are consecutive to each other. Consecutive means side by side or one day ahead or behind.
Favorable outcomes are: (T,W), (W,TH), (TH,F), (F,S), (W,T),(TH,W), (F,TH), (S,F)
So the number of favorable outcomes is 8.
We use the formula for probability of an event which is given by the number of favorable outcomes divided by the total number of outcomes.
Probability that Shyam and ekta visit shop on consecutive days \[ = \dfrac{8}{{25}}\]

(iii) Different days:
Since we know the probability of Shyam and Ekta visiting the shop on the same day is \[\dfrac{1}{5}\].
So, the probability of Shyam and Ekta visiting the shop on different days will be \[1 - \]probability of Shyam and Ekta visiting the shop on the same day.
So, So, probability of Shyam and Ekta visiting the shop on different days \[ = 1 - \dfrac{1}{5}\]
Taking LCM we can write
Probability \[ = \dfrac{{5 - 1}}{5}\]
Solving the fraction we get
Probability of Shyam and Ekta visiting the shop on different days \[ = \dfrac{4}{5}\].
Note:Students can many times make the mistake of leaving out some of the total outcomes when they try to write the outcomes roughly, they are advised to make use of the table to write the outcomes which later helps in selection of favorable outcomes as well.Also, keep in mind the answer of probability should always be greater than or equal to zero and less than or equal to one.