Lesson Twisted probability problems

Twisted probability problems

Problem 1

Let G be the number of green hats and B be the number of blue hat.


The probability that two randomly chosen hats are both green is  P(GG) = .

The probability that two randomly chosen hats are both blue  is  P(BB) = .


The probability that two randomly chosen hats are the same color is  

    P(GG) + P(BB) =  + .



The probability that two randomly chosen hats are of different color is  

    P(GB) + P(BG) =  +  = .


So, your first equation, from the condition, is

    P(GG) + P(BB) = P(GB) + P(BG),  or   +  = .


Canceling common denominators, you get this equation in equivalent form

    G*(G-1) + B*(B-1) = 2GB.       (1)


We completed with the first part of the condition, and now start working with the second part.


From the second part, you have this equation

    P(GG) = P(B),  or   = .


After canceling common factors in the denominators, it takes the form

     = B, or, equivalently,  

    G*(G-1) = B*(G + B - 1)      (2)


So, now our task is to solve the system of equations (1) and (2).

For it, replace  G*(G-1) in the left side of (1) by  B*(G+B-1),  based on (2).  You will get then

    B*(G + B - 1) + B*(B-1) = 2GB.


Cancel B in both side

    G + B - 1 + B - 1 = 2G,  or

    2B - 2 = G.                  (3)


Based on (3), replace G in (2) by 2B-2.  You will get

    (2B-2)*(2B - 2 -1) = B*((2B-2) + B -1).


Simplify it step by step

    (2B-2)*(2B-3) = B*(3B-3)

    4B^2 - 4B - 6B + 6 = 3B^2 - 3B

    B^2 - 7B + 6 = 0

    (B-6)*(B+1) = 0


Only positive root  B = 6  makes sense.


ANSWER.  6 blue hats and  2B-2 = 2*6-2 = 10 green hats.  

Problem 2

We are given that EITHER A hits the target first time  at his 5-th shot,

                  OR     B hits the target first time at his 12-th shot,


but we do not know exactly who did it first, A or B.



        Also notice that this "OR" is EXCLUSIVE: either A or B, but not both.



The probability that A hits the target first time with his 5-th shoot  AND  B do not hit target with his 12 shoots is  


    P(A, 5+;  B, 12-) = 0.7*(1-0.7)^4 * (1-0.4)^12 = 0.7*0.3^4 * 0.6^12 = 0.00001234.



The probability that B hits the target first time with his  12-th shoot AND  A do not hit target with his 5 shoots  is


    P(B, 12+;  A, 5-) = 0.4*(1-0.4)^11 * (1-0.7)^5 = 0.4*0.6^11 * 0.3^5 = 0.00000353.

 


The probability that it is A who hits the target first is  


    P(A, 5+;  B, 12-) / ( P(A, 5+;  B, 12-) + P(B, 12+;  A,5-) ) =  = 0.7778.



The probability that it is B who hits the target first is  


    P(B, 12+;  A, 5-) / ( P(A, 5+;  B, 12-) + P(B, 12+;  A,5-) ) =  = 0.2222.



ANSWER.  The probability that A hits the target first is  0.7778.

         The probability that B hits the target first is  0.2222,