Detective X, Need a Promotion?

Probability Level 3

Detective X has had two criminals J and K under surveillance for a long time, and has found the following pattern regarding their secret meetings:

  • They show up at their secret place in the early afternoon, not every single day, but J with probability 3 4 \frac34 and K with 2 3 . \frac23.
  • If any of them comes, he is equally likely to show up at any time between noon and 2:00 pm, independent of the other.
  • On the day only one of them comes, the man showing up waits for the other until 2:00 pm.
  • If they both show up, together they leave the place--after secret talks--half way between the exact time of rendezvous and 2:00 pm.

Today, which is the last day of their meeting according to a source, detective X finally gets the permission to arrest them from his police headquarters. He is so excited because he will certainly get a promotion if he catches both these criminals today. Unfortunately, however, because of heavy traffic he arrives at the secret place a bit too late, at 1:30 pm, and there is nobody there! Had he arrived before noon, the chance of getting them both (and a promotion) would have been 3 4 × 2 3 = 1 2 = \frac34 \times \frac23=\frac12= 50%.

But, right now, what's the approximate chance of detective X getting a promotion?

3% 9% 17% 25% 50%

Jimin Khim by Jimin Khim

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1 solution

Jimin Khim Staff
Oct 1, 2017

Finding neither of them there at 1:30 pm can only happen in one of the following 5 cases (with their respective probabilities in parentheses):

  1. Today is one of those days neither of them shows up. ( 1 4 × 1 3 = 1 12 = 8 96 ) \left(\frac14 \times \frac13=\frac1{12}=\frac{8}{96} \right)
  2. J will not show up today, but K will show up any time before 2:00 pm. ( 1 4 × ( 2 3 × 30 120 ) = 1 24 = 4 96 ) \left(\frac14 \times \big(\frac23 \times \frac{30}{120}\big)=\frac1{24}=\frac{4}{96} \right)
  3. K will not show up today, but J will show up any time before 2:00 pm. ( ( 3 4 × 30 120 ) × 1 3 = 1 16 = 6 96 ) \left(\big(\frac34 \times \frac{30}{120}\big) \times \frac13=\frac1{16}=\frac{6}{96} \right)
  4. Both will show up any time before 2:00 pm. ( ( 3 4 × 30 120 ) × ( 2 3 × 30 120 ) = 1 32 = 3 96 ) \left(\big(\frac34 \times \frac{30}{120}\big) \times \big(\frac23\times\frac{30}{120}\big) =\frac1{32}=\frac{3}{96} \right)
  5. They already met up with each other and left the place. ( ( 3 4 × 60 120 ) × ( 2 3 × 60 120 ) = 1 8 = 12 96 ) \left(\big(\frac34 \times \frac{60}{120}\big) \times \big(\frac23\times\frac{60}{120}\big) =\frac1{8}=\frac{12}{96} \right)

In Case 5, note that if they both had shown up between noon and 1:00 pm, they would definitely be gone by 1:30 pm because 1:30 pm is exactly half way between 1:00 pm and 2:00 pm. (If, for example, their rendezvous time was 1:10 pm, then they both would still be there because their departure time would be 1:35 pm.) Therefore, for detective X to be able to arrest them both today, Case 4 must be happening, the probability of which is 3 96 8 96 + 4 96 + 6 96 + 3 96 + 12 96 = 3 33 = 1 11 ≈ 9 % . □ \frac{\frac{3}{96}}{\frac{8}{96}+\frac{4}{96}+\frac{6}{96}+\frac{3}{96}+\frac{12}{96}}=\frac{3}{33}=\frac{1}{11}\approx 9\%.\ _\square

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Wow. This is really good.

Rui Chang Lu - 3 years, 8 months ago

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