probability that in a group of three friends no two of them
- Thread starter brucejin
- Start date
Hello, brucejin!
\(\displaystyle \text{The first friend can have any weekday for his birthday: }\:\frac{7}{7}\)
\(\displaystyle \text{The second friend must not have the the first one's birthday: }\:\frac{6}{7}\)
\(\displaystyle \text{The third friend must not have the first two birthdays: }\:\frac{5}{7}\)
\(\displaystyle \text{Therefore: }\
(\text{3 different days}) \;=\;\frac{7}{7}\cdot\frac{6}{7}\cdot\frac{5}{7} \;=\;\frac{30}{49}\)
\(\displaystyle \text{The first friend can have any weekday for his birthday: }\:\frac{7}{7}\)
\(\displaystyle \text{The second friend must not have the the first one's birthday: }\:\frac{6}{7}\)
\(\displaystyle \text{The third friend must not have the first two birthdays: }\:\frac{5}{7}\)
\(\displaystyle \text{Therefore: }\
(\text{3 different days}) \;=\;\frac{7}{7}\cdot\frac{6}{7}\cdot\frac{5}{7} \;=\;\frac{30}{49}\)
D
Deleted member 4993
Guest
brucejin said:
One way to look at it would be:
The first person can be born on any of the 3 days. Prob. = 3/7
The second person can be born on any of the rest of the 2 days. Prob. = 2/7
The third person can be born on the day (one) left . Prob. = 1/7
Thanks.
3/7 x 2/7 x 1/7 is cool.
More questions
(1) "What is the probability that in a group of three friends all born on Monday?"
Is the answer 1/7 x 1/7 x 1/7 = 1/343?
(2) "What is the probability that in a group of three friends all born on the same weekday?"
Is the answer 7/7 x 1/7 x 1/7 = 1/49?
3/7 x 2/7 x 1/7 is cool.
More questions
(1) "What is the probability that in a group of three friends all born on Monday?"
Is the answer 1/7 x 1/7 x 1/7 = 1/343?
(2) "What is the probability that in a group of three friends all born on the same weekday?"
Is the answer 7/7 x 1/7 x 1/7 = 1/49?
D
Deleted member 4993
Guest
brucejin said:
Looks good to me.
D
Deleted member 4993
Guest
brucejin said:
It will help you a lot - if you reasoned out the answer given by pka!
If the numbers were to become large - suppose you are doing the problem for dates in June and for 15 students - you will spend awful lot of time trying to figure out any other way (and probably that is the type of problem they would ask in the finals - at least I would).