Lesson Probability problems similar to a coinciding birthdays problem

Probability problems similar to a coinciding birthdays problem

Problem

Consider 7 people A, B, C, D, E, F and G.

Person A has birthday in some of 12 months, X.

The probability that person B has birthday in some other month Y, different from X, is  .

The probability that person C has birthday in some other month Z, different from X and Y, is  .

The probability that person D has birthday in some other month W, different from X, Y and Z, is  .


Thinking this way, we get that the probability for 7 persons to have birthdays in different months is

    ,

which is the product of 6 factors; the structure of the formula is clear.


Making calculations, you get value  P = 0.111400463, or 0.11 = 11% rounded.


So, the probability that 7 randomly chosen people have birthdays in different months is 0.11, or 11%.


The probability that of 7 randomly chosen people at least two of them have 
birthdays in the same month is the complement to 0.11, i.e. 1-0.11 = 0.89.


Thus the probability that John   wins is 0.89, or 89%; 
     the probability that Krista wins is 0.11, or 11%.


ANSWER.  The probability that John wins is 0.89 = 89%.

         The probability that Krista wins is 0.11 = 11%.

Problem 2

We have 26 lowercase letters and 26 uppercase letters in English alphabet.


Assuming that we use lowercase letters only, there are   possible words 
of the length 11.


Of them, the number of words with no repeating letters is  26*25*24*23*22*21*20*19*18*17*16
(11 factors of integer numbers in descending order starting from 26, in all).


The probability do not have repeating letters is


    1 -  = 


    = 1 - 0.084026 = 0.915974   (rounded).    ANSWER

Problem 3

For 10 passengers, make all possible lists/schedules, at which of 14 floors above
they can get off.


1st passenger can get off at any of 14 floors above, giving 14 options.
2nd passenger can get off at any of 14 floors above, giving 14 options.
3rd passenger can get off at any of 14 floors above, giving 14 options.
. . . . . . . and so on . . . . . . . 
10th passenger can get off at any of 14 floors above, giving 14 options.


So, in all, there are    different possible schedules.


Next, let's calculate, how many of these schedules have different floors for different people.


1st passenger can get off at any of 14 floors above, giving 14 options.
2nd passenger can get off at any of remaining 13 floors, giving 13 options.
3rd passenger can get off at any of remaining 12 floors, giving 12 options.
. . . . . . . and so on . . . . . . . 
10th passenger can get off at any of remaining 14-9 = 5 floors, giving 5 options.


So, the number of schedules that have have different floors for different people is  14*13*12*. . . *5.


Thus, the probability that no two passengers get out on some floor is

     = 0.01256  (rounded).


From here, the probability that at least two of these passengers will get off 
at the same floor is the COMPLEMENT of this value to 1

    P = 1 -  = 1 - 0.01256 = 0.98744 (rounded).


ANSWER.  The probability that at least two of these passengers will get off at the same floor is

             P = 1 -  = 1 - 0.01256 = 0.98744 (rounded)

Problem 4

There are 20 separate pieces in the box (of 10 pairs of gloves).


The probability the problem asking for, is the ratio, whose denominator is the number of all 
sextuples that can be selected from the set of 20 items,   = 38760.


The numerator is the number of all sextuples that have one matching pair and 4 non-matching items.
Our task is to calculate the number of such sextuples.


In turn, this number is the product of 10 (the number of gloves) multiplied by the number 
of all possible quadruples that consist of non-matching pairs from remaining 2*(10-1) = 18 items.


Thus our task is to calculate the number of all non-matching quadruples of 18 separate pieces.


We can select 1st item as any of 18 pieces, which gives us 18 options.

After that, we can choose next item from remaining 18-2 = 16 items (excluding item #1 and its twin).
      It gives us 16 possible options.

After that, we can choose 3rd item from remaining 16-2 = 14 items (excluding item #1 and its twin,
      as well as item #2 and its twin). It gives us 14 possible options.

After that, we can choose 4th item from remaining 14-2 = 12 items (excluding item #1 and its twin,
      item #2 and its twin, item #3 and its twin). It gives us 12 possible options.


So, the number of all different quadruples of non-matching items from the total of 18 items (9 gloves) 
      is 18*16*14*12 = 48384.


          Notice that 48384 is the number of all such different    quadruples, 
          so we need to divide this number by 4! = 24, to count    quadruples.


Thus, the number of unordered quadruples is  48384/24 = 2016,
and   the number of all sextuples we are interesting to count is 10*2016 = 20160.


Now the probability we are looking for is  P =  =  = 0.5201  (rounded).     ANSWER