Question

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. What is the probability that both will visit the shop on different days?

OR

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

[3 Marks]

Answer 1

Total favourable outcomes associated to the random experiment of visiting a particular shop in the same week (Tuesday to Saturday) by two customers Shyam and Ekta are:
(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)
$\therefore$ Total number of outcomes = 25
(1 Mark)

Number of favourable outcomes of visiting on different days are 25 - 5 = 20
$\therefore$ Number of favourable outcomes = 20
(1 Mark)

Hence, required probability = $\frac{20}{25}=\frac{4}{5}$
(1 Mark)

OR

Let the unit digit and tens digits of the number be x and y respectively.
Then, the number will be $10y+x$
Number after reversing the digits is $10x+y$
According to the question,
$x+y=9...\left(i\right)$
$9(10y+x)=2(10x+y)$
$88y-11x=0$
$-x+8y=0...\left(ii\right)$
[1 Mark]
Adding equation (i) and (ii), we get
$9y=9$
$y=1...\left(iii\right)$
Putting the value in equation (i), we get
$x=8$
​​​​​​​ [1 Mark]

Hence, the number is $10y+x=10\times 1+8=18$
[1 Mark]