# Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. Find the probability that both will visit the shop on

(a) the same day?

(b) consecutive days?

(c) different days?

Answer

Verified

362.1k+ views

Hint: While solving this problem of probability, we start with finding the total possible number of outcomes in this problem. The number of days from Tuesday to Saturday (including Tuesday and Saturday) is 5. Further, we calculate the outcome for two people (Shyam and Ekta), the total outcomes are $5\times 5=25$.

Complete step-by-step answer:

To make this clear, the possible outcomes are listed below

(T, T) (T, W) (T, Th) (T, F) (T, S)

(W, T) (W, W) (W, Th) (W, F) (W, S)

(Th, T) (Th, W) (Th, Th) (Th, F) (Th, S)

(F, T) (F, W) (F, Th) (F, F) (F, S)

(S, T) (S, W) (S, Th) (S, F) (S, S)

Where,

T stands for Tuesday

W stands for Wednesday

Th stands for Thursday

F stands for Friday

S stands for Saturday

We can clearly see that the listed outcomes are 25.

(a) the same days

From the list of outcomes, we can clearly see that for Ekta and Shyam to visit the particular shop on the same day, the desirable number of outcomes are 5. These are (T, T), (W, W), (Th, Th), (F, F), (S, S). Now,

$\begin{align}

& \text{probability}=\dfrac{\text{desired number of outcomes}}{\text{total number of outcomes}} \\

& \text{probability}=\text{ }\dfrac{5}{25} \\

& \text{probability}=\text{ }\dfrac{1}{5} \\

\end{align}$

(b) consecutive days

From the list of outcomes, we can clearly see that for Ekta and Shyam to visit the particular shop on consecutive days, the desirable number of outcomes are 8. These are (T, W), (W, T), (W, Th), (Th, W), (Th, F), (F, Th), (F, S), (S, F).

$\begin{align}

& \text{probability}=\dfrac{\text{desired number of outcomes}}{\text{total number of outcomes}} \\

& \text{probability}=\text{ }\dfrac{8}{25} \\

\end{align}$

(c) different days

To solve this, we simply need to subtract the probability for Ekta and Shyam to visit on the same days from 1. To explain this,

Probability (same days) + Probability (different days) = 1

Since, when Ekta and Shyam don’t visit a particular shop on the same day, they visit the particular shop on different days. Thus, the sum of their probabilities should add up to 1.

\[\begin{align}

& \text{probability}=\text{ }1-\dfrac{1}{5} \\

& \text{probability}=\text{ }\dfrac{4}{5} \\

\end{align}\]

Note: A common mistake while calculating the probability for Ekta and Shyam to visit on consecutive days is to forget in counting all the desirable number of outcomes. To explain, one can forget to count (T, W), (W, T) as two separate outcomes and thus land up with the desired number of outcomes as 4. (namely, (T, W), (W, Th), (Th, F), (F, S)). This may lead to the calculation of incorrect probability.

Complete step-by-step answer:

To make this clear, the possible outcomes are listed below

(T, T) (T, W) (T, Th) (T, F) (T, S)

(W, T) (W, W) (W, Th) (W, F) (W, S)

(Th, T) (Th, W) (Th, Th) (Th, F) (Th, S)

(F, T) (F, W) (F, Th) (F, F) (F, S)

(S, T) (S, W) (S, Th) (S, F) (S, S)

Where,

T stands for Tuesday

W stands for Wednesday

Th stands for Thursday

F stands for Friday

S stands for Saturday

We can clearly see that the listed outcomes are 25.

(a) the same days

From the list of outcomes, we can clearly see that for Ekta and Shyam to visit the particular shop on the same day, the desirable number of outcomes are 5. These are (T, T), (W, W), (Th, Th), (F, F), (S, S). Now,

$\begin{align}

& \text{probability}=\dfrac{\text{desired number of outcomes}}{\text{total number of outcomes}} \\

& \text{probability}=\text{ }\dfrac{5}{25} \\

& \text{probability}=\text{ }\dfrac{1}{5} \\

\end{align}$

(b) consecutive days

From the list of outcomes, we can clearly see that for Ekta and Shyam to visit the particular shop on consecutive days, the desirable number of outcomes are 8. These are (T, W), (W, T), (W, Th), (Th, W), (Th, F), (F, Th), (F, S), (S, F).

$\begin{align}

& \text{probability}=\dfrac{\text{desired number of outcomes}}{\text{total number of outcomes}} \\

& \text{probability}=\text{ }\dfrac{8}{25} \\

\end{align}$

(c) different days

To solve this, we simply need to subtract the probability for Ekta and Shyam to visit on the same days from 1. To explain this,

Probability (same days) + Probability (different days) = 1

Since, when Ekta and Shyam don’t visit a particular shop on the same day, they visit the particular shop on different days. Thus, the sum of their probabilities should add up to 1.

\[\begin{align}

& \text{probability}=\text{ }1-\dfrac{1}{5} \\

& \text{probability}=\text{ }\dfrac{4}{5} \\

\end{align}\]

Note: A common mistake while calculating the probability for Ekta and Shyam to visit on consecutive days is to forget in counting all the desirable number of outcomes. To explain, one can forget to count (T, W), (W, T) as two separate outcomes and thus land up with the desired number of outcomes as 4. (namely, (T, W), (W, Th), (Th, F), (F, S)). This may lead to the calculation of incorrect probability.

Last updated date: 26th Sep 2023

•

Total views: 362.1k

•

Views today: 4.62k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE