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September 10

Friends meeting at the park

Three people come and meet at the park each day. Each comes 7 out of 10 days. What is the chance of all three coming? Only two? Only one? None? Thanks. I'm stumped. Please try to answer in a way that a complete idiot (me) will understand. Actually, the reason isn't so imporant. It is the actual % chance that I'm after. Many many thanks. :) Anna Frodesiak (talk) 03:32, 10 September 2014 (UTC)[reply]

We'd have to start with the assumption that each person showing up is an independent event. This probably isn't correct, as they may all avoid rainy days, or all show up when they planned to meet. But, if we assume each has an independent chance of showing up 70% of the time, then the chances of all or none showing up are:

Zero: 0.3³ = 0.027
Three: 0.7³ = 0.343

Now the chances of 1 or 2 showing up are complicated by the fact that a different one or two might show up, so we have to account for all the ways that can happen. In this case there are 3 ways 1 person can show up (A, B or C) and 3 ways 2 people can show up (A&B, A&C, or B&C):

One: 3(0.3²0.7¹) = 0.189
Two: 3(0.3¹0.7²) = 0.441

To check our work, add them all up and you should get 1.0, or, if we multiply all the numbers by 100, we get percentages: 2.7% + 34.3% + 18.9% + 44.1% = 100%. StuRat (talk) 04:42, 10 September 2014 (UTC)[reply]

Wow! You are a super-genius. I am very impressed. I sort of figured out the zero and three part, but got stuck on the one and two. A thousand thanks for your help. :) :) :) Yay StuRat! And yay refdesk. The best kept secret on the Internet. :) Anna Frodesiak (talk) 07:37, 10 September 2014 (UTC)[reply]

You're quite welcome. Here it is, presented in the tree diagram format mentioned below (or as close as I can get using ASCII text):

                            I N D E P E N D E N T   E V E N T S
              +-----------------------------------------------------------------------+
    Person A: |     P R E S E N T   ( 0 . 7 )     |      A B S E N T   ( 0 . 3 )      |
              +-----------------+-----------------+-----------------+-----------------+   
    Person B: |  Present (0.7)  |   Absent (0.3)  |  Present (0.7)  |   Absent (0.3)  |
              +--------+--------+--------+--------+--------+--------+--------+--------+
    Person C: | P (0.7)| A (0.3)| P (0.7)| A (0.3)| P (0.7)| A (0.3)| P (0.7)| A (0.3)|
              +--------+--------+--------+--------+--------+--------+--------+--------+
            / |.7×.7×.7|.7×.7×.3|.7×.3×.7|.7×.3×.3|.3×.7×.7|.3×.7×.3|.3×.3×.7|.3×.3×.3|
Probability   +--------+--------+--------+--------+--------+--------+--------+--------+
            \ | 0.343  | 0.147  | 0.147  | 0.063  | 0.147  | 0.063  | 0.063  | 0.027  |
              +--------+--------+--------+--------+--------+--------+--------+--------+
   # Present: |    3   |    2   |    2   |    1   |    2   |    1   |    1   |    0   |
              +--------+--------+--------+--------+--------+--------+--------+--------+
 
 3 present = 0.343                         = 34.3%
 2 present = 0.147 + 0.147 + 0.147 = 0.441 = 44.1%
 1 present = 0.063 + 0.063 + 0.063 = 0.189 = 18.9% 
 0 present = 0.027                         =  2.7%
                                            ------
                                            100.0%

Note that tree diagrams are only practical for a small number of events, with a small number of possible outcomes for each event. Here we have 3 events, with two outcomes each (3 people who can be present or absent), making for 2³ or 8 possible outcomes. If we had 10 events with 2 outcomes each, that would give us 1024 possible outcomes, or if we had 3 events with 10 possible outcomes each, that would give us 1000 possible outcomes. Either would be way too big to draw as a tree. But, if you can draw a tree, it can help to visualize dependencies on events, as well as if all events are independent. For example, let's say person A and B are a couple, and always are present (0.7) or absent (0.3) at the same time. The presence of person C (0.7) remains an independent event:

                                D E P E N D E N T   E V E N T S   ( A = B )
              +-----------------------------------------------------------------------+
    Person A: |     P R E S E N T   ( 0 . 7 )     |      A B S E N T   ( 0 . 3 )      |
              +-----------------+-----------------+-----------------+-----------------+   
    Person B: |  Present (1.0)  |   Absent (0.0)  |  Present (0.0)  |   Absent (1.0)  |
              +--------+--------+--------+--------+--------+--------+--------+--------+
    Person C: | P (0.7)| A (0.3)| P (0.7)| A (0.3)| P (0.7)| A (0.3)| P (0.7)| A (0.3)|
              +--------+--------+--------+--------+--------+--------+--------+--------+
            / |.7×1×.7 |.7×1×.3 |.7×0×.7 |.7×0×.3 |.3×0×.7 |.3×0×.3 |.3×1×.7 |.3×1×.3 |
Probability   +--------+--------+--------+--------+--------+--------+--------+--------+
            \ |  0.49  |  0.21  |    0   |    0   |    0   |    0   |  0.21  |  0.09  |
              +--------+--------+--------+--------+--------+--------+--------+--------+
   # Present: |    3   |    2   |    2   |    1   |    2   |    1   |    1   |    0   |
              +--------+--------+--------+--------+--------+--------+--------+--------+
 
 3 present = 0.49             = 49%
 2 present = 0.21 + 0 + 0     = 21%
 1 present = 0    + 0 + 0.21  = 21% 
 0 present = 0.09             =  9%
                               ----
                               100%

StuRat (talk) 13:58, 10 September 2014 (UTC)[reply]

You should also take a look at Binomial distribution. -- Meni Rosenfeld (talk) 09:05, 10 September 2014 (UTC)[reply]

You mean me? That page could be upside down and scrambled and would make as much sense to me. Thank you, though. :) Anna Frodesiak (talk) 09:16, 10 September 2014 (UTC)[reply]

Yeah, I've noticed that page isn't very newbie-friendly. But it describes the general way to solve problems like the one you've presented. If there are n different things which can either happen or not, each with probability p, and they are independent, then the probability that exactly k of them will happen is

{\frac {n!}{k!(n-k)!}}p^{k}(1-p)^{n-k}

, where n! is the factorial. In your case, the things that happen are each person showing up,

n=3

and

p=0.7

. -- Meni Rosenfeld (talk) 12:49, 10 September 2014 (UTC)[reply]

A good way to understand this sort of thing "visually" is with a tree diagram. That article is just a stub, but the linked BBC page has some good examples. If you take their 3-coin-tosses example, and replace the 3 tosses with the arrival or non-arrival of each of the 3 people (and change the 0.5 probabilities of heads and tails to 0.7 and 0.3) then you should end up with the same answers as above. AndrewWTaylor (talk) 13:22, 10 September 2014 (UTC)[reply]

Holy moly. I'm actually understanding this. I sort of lost it half way through, but was getting it. I will read it again tomorrow after a big coffee. This is very nice. I never understand stuff like this. I have the IQ of lichen. Anna Frodesiak (talk) 14:47, 10 September 2014 (UTC)[reply]

Awesome, glad we could help. StuRat (talk) 16:57, 10 September 2014 (UTC)[reply]