# 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 anotherday. What is the probability that both will visit the shop oni) the same day?ii) consecutive day?iii) Different day?

total no. of elementary events = 25

because they can visit the shop on 2 days in 5 * 5 days

1) tuesday,      wednesday        thursday,             friday,         saturday

favourable no. of ways = 6

probability that they will visit the shop on same day = 6/36 = 1/6

2)

favourable outcomes are:

tues,wed    wed,thurs    thurs,fri    fri,sat

probability that they visit on consecutive days = 4/36 = 1/9

3)

probability that they visit on different days = 1 - probability that they visit on same day

1 - 1/6 = 5/6

• -6

cn u plzz elaborate???

• -7
 tues wed thurs fri sat tues tues,tues tues,wed tues,thurs tues,fri tues,sat wed wed,tues wed,wed wed,thurs wed,fri wed.sat thurs thurs,tues thurs,wed thurs,thurs thurs,fri thurs,sat fri fri,tues fri,wed fri,thurs fri,fri fri,sat sat sat,tues sat,wed sat,thurs sat,fri sat,sat

first row and first coloumn represent shyam and ekta visiting the shop

the rest is the result

this problem is similiar as how we do the problem of finding probability we 2 dice are thrown

total no. of elementary events = 25

because they can visit the shop on 2 days in 5 * 5 ways

1) tuesday, wednesday thursday, friday, saturday

favourable no. of ways = 6

probability that they will visit the shop on same day = 6/36 = 1/6

2)

favourable outcomes are:

tues,wed wed,thurs thurs,fri fri,sat

probability that they visit on consecutive days = 4/36 = 1/9

3)

probability that they visit on different days = 1 - probability that they visit on same day

1 - 1/6 = 5/6

• -2

 tues wed thurs fri sat tues tues,tues tues,wed tues,thurs tues,fri tues,sat wed wed,tues wed,wed wed,thurs wed,fri wed.sat thurs thurs,tues thurs,wed thurs,thurs thurs,fri thurs,sat fri fri,tues fri,wed fri,thurs fri,fri fri,sat sat sat,tues sat,wed sat,thurs sat,fri sat,sat

first row and first coloumn represent shyam and ekta visiting the shop

the rest is the result

this problem is similiar as how we do the problem of finding probability we 2 dice are thrown

total no. of elementary events = 25

because they can visit the shop on 2 days in 5 * 5 ways

1) tuesday, wednesday thursday, friday, saturday

favourable no. of ways = 5

probability that they will visit the shop on same day = 5/2536 = 1/5

2)

favourable outcomes are:

tues,wed wed,thurs thurs,fri fri,sat

probability that they visit on consecutive days = 4/25

3)

probability that they visit on different days = 1 - probability that they visit on same day

1 - 156 = 4/5

• 12

in the last part of the question, probability = 1 - 1/5 = 4/5

• 2

for visiting in consecutive days (ii)there are 8 favourable out comes not just 4,i.eshyam S on tue ekta E on wedekta E on tue shyam S on wedlike that vice versa

so probability is 8/25.

• 3
Solution:The total number of days is 5 and hence both of them can reach the shop in 5 ways.Hence, total number of outcomes = 5 x 5 = 25 They can reach on the same day in 5 ways, i.e. (Tue Tue), (Wed Wed), (Thur Thur), (Fri Fri) and (Sat Sat) P(Reaching on same day)10 probability exercise solution They can reach on consecutive days in following 8 ways: (tue wed), (wed, tue), (wed thur), (thur wed), (thur fri), (fri thu), (fri sat), (sat fri) P(Reaching on consecutive days)10 probability exercise solution Since P (reaching on same days) = 1/5 Hence, P(reaching on different days)10 probability exercise solution.
• 35
Total number of days = 5. So Shyam and Ekta may visit the shop in 5 ways So total number of outcomes = 5*5 = 25. 1. They can reach on the same day in 5 ways, i.e. (Tuesday-Tuesday), (Wednesday-Wednesday), (Thursday-Thursday), (Friday-Friday) and (Saturday-Saturday) So, probability that both will visit the shop on the same day = 5/25 = 1/5. 2. They can reach on consecutive days in following 8 ways: (tuesday-wednesday), (wednesday- tuesday), (wednesday-thursday), (thursday- wednesday), (thursday- friday), (friday-thusday), (friday-saturday), (saturday-friday). probability that both will visit the shop on the consecutive day = 8/25. 3. probability that both will visit the shop on the same day = 1/5 So probability that both will visit the shop on the different day =1 - 1/5 = 4/5
• 3
Solution:The total number of days is 5 and hence both of them can reach the shop in 5 ways.Hence, total number of outcomes = 5 x 5 = 25 They can reach on the same day in 5 ways, i.e. (Tue Tue), (Wed Wed), (Thur Thur), (Fri Fri) and (Sat Sat) P(Reaching on same day)10 probability exercise solution They can reach on consecutive days in following 8 ways: (tue wed), (wed, tue), (wed thur), (thur wed), (thur fri), (fri thu), (fri sat), (sat fri) P(Reaching on consecutive days)10 probability exercise solution Since P (reaching on same days) = 1/5 Hence, P(reaching on different days)10 probability exercise solution.
• 2
You all are wring
• 1
You all are wrong
• 0
the question is too long to answer sorry
• 0
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