Hello, psoft!
What is the level of this course?
\(\displaystyle \;\;\)Where did this problem come from?
Is that the original wording of the problem?
\(\displaystyle \;\;\)If so, it is badly written.
What is the probability of two people arriving at a restaurant within 6 minutes of each other between 1 PM and 5 PM?
Are we to assume that A and B will arrive at times rounded to the nearest
minute?
\(\displaystyle \;\;\)If their arrivals are timed to the nearest
second, the problem is changed drastically.
Does "<u>within</u> 6 minutes" means 0, 1, 2, 3, 4, or 5 minutes between arrivals?
Does "<u>between</u> 1 PM and 5 PM" exclude the endpoints?
Assuming "nearest minute",
\(\displaystyle \;\;\)and 0 to 5 minutes between arrivals,
\(\displaystyle \;\;\)and literally <u>between</u> 1:00 and 5:00,
I have a soltuion . . . but it's long and tedious.
Number the times from 1 to 239.
\(\displaystyle \;\;1:01 = 1,\:1:02 = 2,\:1:03\,=\,3,\,\cdots,\,4:59\,=\,239\)
Suppose A arrives at \(\displaystyle \,t\,=\,1.\;\;\)Denote this as \(\displaystyle A_1\)
Let \(\displaystyle B\) means "B arrives within 6 minutes of \(\displaystyle A\)."
Now we will consider all 239 possible arrivals for \(\displaystyle A\).
\(\displaystyle P(A_1)\,=\,\frac{1}{239}\)
Then B can arrive at: 1, 2, 3, 4, 5, 6 . . . 6 choices. \(\displaystyle \;P(B)\,=\,\frac{6}{240}\)
Hence: \(\displaystyle \,P(A_1\,\cap\,B)\,=\,\left(\frac{1}{240}\right)\left(\frac{6}{240}\right)\,=\,\frac{6}{240^2}\)
\(\displaystyle P(A_2)\,=\,\frac{1}{240}\)
Then B can arrive at: 1, 2, 3, 4, 5, 6, 7 . . . 7 choices. \(\displaystyle \;P(B)\,=\,\frac{7}{240}\)
Hence: \(\displaystyle \,P(A_2\,\cap\,B)\,=\,\left(\frac{1}{240}\right)\left(\frac{7}{240}\right)\,=\,\frac{7}{240^2}\)
\(\displaystyle P(A_3)\,=\,\frac{1}{240}\)
Then B can arrive at: 1, 2, 3, 4, 5, 6, 7, 8 . . . 8 choices: \(\displaystyle \;{(B)\,=\,\frac{8}{240}\)
Hence: \(\displaystyle \,P(A_3\,\cap\,B)\,=\,\left(\frac{1}{240}\right)\left(\frac{8}{240}\right)\,=\,\frac{8}{240^2}\)
\(\displaystyle P(A_4)\,=\,\frac{1}{240}\)
Then B can arrive at: 1, 2, 3, 4, 5, 6, 7, 8, 9 . . . 9 choices: \(\displaystyle \;P(B)\,=\,\frac{9}{240}\)
Hence: \(\displaystyle \,P(A_4\,\cap\,B)\,=\,\left(\frac{1}{240}\right)\left(\frac{9}{240}\right)\,=\,\frac{9}{240^2}\)
\(\displaystyle P(A_5)\,=\,\frac{1}{240}\)
Then B can arrive at: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 . . . 10 choices: \(\displaystyle \;P(B)\,=\,\frac{10}{240}\)
Hence: \(\displaystyle \;P(A_5\,\cap\,B)\,=\,\left(\frac{1}{240}\right)\left(\frac{10}{240}\right)\,=\,\frac{10}{240^2}\)
\(\displaystyle P(A_6)\,=\,\frac{1}{240}\)
Then B can arrive at: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 . . . 11 choices: \(\displaystyle \;P(B)\,=\,\frac{11}{240}\)
Hence: \(\displaystyle \;P(A_6\,\cap\,B)\,=\,\left(\frac{1}{240}\right)\left(\frac{11}{240}\right)\,=\,\frac{11}{240^2}\)
This probability is the same for \(\displaystyle A_6\) to \(\displaystyle A_{234}\) . . . 229 cases.
These probabilities will total: \(\displaystyle \,229\left(\frac{1}{240}\right)\left(\frac{11}{240}\right)\,=\,\frac{229(11)}{240^2}\)
\(\displaystyle P(A_{235)\,=\,\frac{1}{240}\)
Then B can arrive at: 230, 231, 232, 233, 234, 235, 236, 237, 238, 239 . . . 10 cases: \(\displaystyle \
(B)\,=\,\frac{10}{240}\)
Hence: \(\displaystyle \;P(A_{235}\,\cap\,B)\,=\,\left(\frac{1}{240}\right)\left(\frac{10}{240}\right)\,=\,\frac{10}{240^2}\)
\(\displaystyle P(A_{236})\,=\,\frac{1}{240}\)
Then B can arrive at: 231, 232, 233, 234, 235, 236, 237, 238, 239 . . .9 choices: \(\displaystyle \;P(B)\,=\,\frac{9}{240}\)
Hence: \(\displaystyle \;P(A_{236}\,\cap\,B)\,=\,\left(\frac{1}{240}\right)\left(\frac{9}{240}\right)\,=\,\frac{9}{240^2}\)
\(\displaystyle P(A_{237})\,=\,\frac{1}{240}\)
Then B can arrive at: 232, 233, 234, 235, 236, 237, 238, 239 . . . 8 choices: \(\displaystyle \;P(B)\,=\,\frac{8}{240}\)
Hence: \(\displaystyle \;P(A_{237}\,\cap\,B)\,=\,\left(\frac{1}{240}\right)\left(\frac{8}{240}\right)\,=\,\frac{8}{240^2}\)
\(\displaystyle P(A_{238})\,=\,\frac{1}{240}\)
Then B can arrive at: 233, 234, 235, 236, 237, 238, 239 . . . 7 choices: \(\displaystyle \;P(B)\,=\,\frac{7}{240}\)
Hence: \(\displaystyle \;P(A_{238}\,\cap\,B)\,=\,\left(\frac{1}{240}\right)\left(\frac{7}{240}\right)\,=\,\frac{7}{240^2}\)
\(\displaystyle P(A_{239})\,=\,\frac{1}{240}\)
Then B can arrive at: 234, 235, 236, 237, 238, 239 . . . 6 choices: \(\displaystyle \;P(B)\,=\,\frac{6}{240}\)
Hence: \(\displaystyle \;P(A_{239}\,\cap\,B)\,=\,\left(\frac{1}{240}\right)\left(\frac{6}{240}\right)\,=\,\frac{6}{240^2}\)
The desired probability is the
sum of these probabilities:
\(\displaystyle \;\;\;\frac{1}{240^2}\,(6\,+\,7\,+\,8\,+\,9\,+\,10\,+\,229\cdot11\,+\,10\,+\,9\,+\,8\,+\,7\,+\,6)\;=\;\frac{2599}{57600}\,=\,0.45121528...\)
Answer: \(\displaystyle \;0.0451\)
