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sh3lsh
 one year ago
What is the probability that in a group of 3 people chosen at random, there are at least two born on the same day of the week?
sh3lsh
 one year ago
What is the probability that in a group of 3 people chosen at random, there are at least two born on the same day of the week?

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sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0Uh, I'll have to parse this slowly. I'll give you the answer tomorrow!

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Alright! Sorry that I can't be of more help!

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Let 1 = sunday 2 = monday 3 = tuesday 4 = wednesday 5 = thursday 6 = friday 7 = saturday So saying a 3 digit string like 173 means "person A was born on sunday, person B was born on saturday, person C was born on tuesday" The entire list would look something like this 111 112 113 114 ... ... ... 774 775 776 777 There are 7^3 = 343 different ways to have the birthdays arranged for the 3 people. This is in terms of the days of the week only. The question is: how many cases arise in which all 3 digits are different? Well we have 3 slots for the 3 people Slot A has 7 choices (1 through 7) Slot B has 6 choices (after picking a digit for slot 1, you have 71 = 6 leftover) Slot C has 5 choices 7*6*5 = 210 So there are 210 ways to have numbers that do not have a repeated digit The other 343  210 = 133 arrangements will have at least one repeat (eg: 113 or 545) So the probability at least 2 were born on the same day of the week is 133/343 = 19/49

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1use the complement, the only way that none of them can have the same birthday is if they all have different days. 7P3, and we already know how many ways there are to present the cases. (1  complement) should do it

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.2Agreed  Find the compliment Chance of 1st Person having a birthday on any weekday is 1/1 Chance of 2nd Person having a birthday different from the 1st is 6/7 Chance that 3rd Person has a birthday different from the 1st two is 5/7 Chance of No birthdays = 1/1 * 6/7 * 5/7 = 30/49

theEric
 one year ago
Best ResponseYou've already chosen the best response.0I'm deleting my responses because I think they're inaccurate and thus might \(\it{hurt}\) someone's understanding.
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