# Lunch Meeting Probability for two person to meet in given 1 hour slot and none would wait more then 15 minute. [closed]

Two friends who have unpredictable lunch hours agree to meet for lunch at their favorite restaurant whenever possible. Neither wishes to eat alone and each dislikes waiting for the other, so they agree that each will arrive at a random time between noon and 1 pm, and each will wait for the other for 15 minutes or until 1:00. What is the probability that the friends will meet for lunch on a given day?

• Do you know what is geometric probability? – Calvin Lin Aug 8 '13 at 15:19
• @DoctorDan - can u post ur comment as answer? – Parimal Raj Aug 8 '13 at 16:47
• 7/16 is one of the option – Parimal Raj Aug 8 '13 at 16:48
• – Henry Sep 15 '16 at 13:25

Per @AppDeveloper's request changing it from a comment to an answer:

Just and idea: consider $0$ to $30$ min, the other half is the same by symmetry.

If A arrives at $0$ min, B has to arrive between $0$ and $15$ min, i.e., $p(B\leq 15|A=0)=\frac{1}{4}$.

If A arrives after $15$ min, $p(B|A)=\frac{1}{2}$.

Applying conditional probabilities and integrating, get for $t \geq 15$ min $p(B|A)p(A)=\frac{1}{8}$ and for $0 \leq t \leq 15$, $p(B|A)p(A)=\frac{3}{32}$ Adding together and multiplying by $2$, get $\frac{7}{16}$.

• "If A arrives at 0 min, B has to arrive between 0 and 51 min, i.e., p(B|A)=1/4."???? Should it be "between 0 to 15"? – user6704 Nov 9 '15 at 17:51
• @Doctor Dan, can you explain in depth how you got for $t \geq 15$ min $p(B|A)p(A)=\frac{1}{8}$ and for $0 \leq t \leq 15$, $p(B|A)p(A)=\frac{3}{32}$ Adding together and multiplying by $2$, get $\frac{7}{16}$? Because aren't you only considering the case where $0 \leq t \leq 15$ – CCC Apr 3 '16 at 22:40

Ah I am surprised the geometric solution has not come up yet. Think of the total times as a square in the coordinate plane, so that each time frame will be 15 mins. The graph will be something like a strip up the diagonal, whose area is very easy to calculate: • Shouldn't the shaded area start at $1/4$ rather than $1/6$? – N. F. Taussig Apr 22 '18 at 18:43
• You are in fact correct! However the description of the geometric solution still stands. – Sultan of Quizikhstan Oct 30 '18 at 16:11

I wanted to verify my solution for this answer, please let me know if my reasoning is correct. Here's how I arrived at $p = \frac{7}{16}$ --

Let $p_1$ be the probability of meeting considering at least one of the witches arrives in $15 \leq t \leq 45$

Witch 1 arrives between 15 min and 45 min mark with probability $p(15 \leq t \leq 45) = \frac{1}{2}$ which gives a thirty min window for witch 2 to arrive, so that their meeting probability is $$p_{1_{w1}} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$

Likewise if we switch witch 1 and 2 $$p_{1_{w2}} = \frac{1}{8}$$

So that $$p_1 = p_{1_{w1}} + p_{1_{w2}} = \frac{1}{4}$$

Let $p_2$ then be the complementary probability where both witches arrive in either $0 \leq t \leq 15$ or $45 \leq t \leq 60$

This will give us overall probability of meeting $p = p_1 + p_2$

$p_2 =$ both witches arrive in time specified $-$ they arrive in exactly opposite time slots i.e. one witch arrives between $0 \leq t \leq 15$ and the other between $45 \leq t \leq 60$

which means $$p_2 = \frac{1}{4} - \frac{1}{16}$$

So that on adding, we get $$p = \frac{1}{4} + \frac{1}{4} - \frac{1}{16}$$

which gives the result $\frac{7}{16}$.

1. No matters who arrives first.
2. Assume $$A$$ arrives first. 2 cases then to be considered: Case 1: $$A$$ arrives between 12.00-12.45. Then the chance that both will meet equals: $$3/4 \cdot 1/4 = 3/16$$. Case 2: A arrives between 12.45-13.00. Then the chance that both will meet equals: $$1/4\cdot 1= 1/4$$.
3. Chance that both will meet then equals: $$3/16 + 1/4 = 7/16$$.