Questions of Probability

Date: 03/22/97 at 18:21:04
From: ac
Subject: Several questions on probability

Question 1. You are shown the following four cards. Each card has a 
single positive integer 1, 2, 3, or 4 on each side, but you can see 
only one side of each card.  Some numbers may appear more than once.  
How many cards must you turn over to verify that any card that has 
a 2 on one side also has a 4 on the opposite side?  

Question 2. A box contains fewer than twenty marbles. If you reach 
into the box and randomly pull out two marbles, without replacing 
them you have a 50 percent chance of getting two blue marbles.  How 
many blue marbles are in the box?

Question 3. Chip and Dale collected thirty-two acorns on Monday and 
stored them with their acorn supply. After Chip fell asleep, Dale ate 
half the acorns. This pattern continued through Friday night, with 
thirty-two acorns being added and half the supply being eaten. On 
Saturday morning, Chip counted the acorns and found that they had only 
thirty-five.  How many acorns had they started with on Monday morning?

Date: 03/22/97 at 19:56:28
From: Doctor Anthony
Subject: Re: Several questions on probability

Question 1. Turn over any card that has either a 2 or a 4 showing on 
it. Under the terms that you specify, none of the four cards need meet 
the condition of a 2 on one side and a 4 on the other side.

Question 2. If there are x blue marbles out of a total of n marbles, 
then the probability of a blue on first draw is  x/n, and the 
probability of a blue on the second draw is (x-1)/(n-1).  

Then we require:

   x/n (x-1)/(n-1) = 1/2
   2x(x-1) = n(n-1).

If n = 18, say, then  

   2x(x-1) = (18)(17)
   x(x-1) = 153
   x^2 - x - 153 = 0
   x = [1 +or- sqrt(1+612)]/2 = 12.879.  

This is not an integer, so n cannot equal 18.

If we take the general case with n marbles, then

   2x^2 - 2x - n(n-1) = 0
   x = [2 + sqrt(4 + 8n(n-1))]/4.

We require 

   4 + 8n(n-1) 

to be a perfect square, or  

   1 + 2n(n-1) 

to be a perfect square, thus

   2n^2 - 2n + 1 

to be a perfect square.

n = 4 gives a value of 25 to this expression.  Then 

   x = [2+2sqrt(25)]/4
     = [2 + 10]/4
     = 12/4
     = 3.

So if we have 4 marbles, 3 of which are blue, we should have a 
probability equal to 1/2 that if we draw out 2 then both will be blue.

Check: (3/4)(2/3) = 2/4 = 1/2.


Question 3. Let n = number they started with on Monday morning.  By 
Monday night they had n + 32.

Tuesday morning   (n+32)/2      Tuesday night (n+32)/2 + 32
Wednesday  am     (n+96)/4      Wednesday pm (n+96)/4 + 32
Thursday   am     (n+224)/8     Thursday  pm (n+224)/8 + 32
Friday     am     (n+480)/16    Friday    pm (n+480)/16 + 32
Saturday   am     (n+992)/32

We require   

   (n+992)/32 = 35
   n+992 = 1120
   n = 128.

So on Monday morning they started with 128 acorns.      

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 03/24/97 at 23:38:30
From: ac
Subject: Several questions on probability

Dr. Anthony,

Thank you for the answers back on my three problems that I sent.  
Also, thank you for putting down the answers.  It really helped me to 
understand it better, by showing the work on the problem and then 
writing down the answer.  I really appreciate your time and 
assistance.
                                                                 
Thanks again,
                                                                            
ac

Date: 04/12/97 at 15:58:22
From: ac
Subject: More questions on probability

Question 4. One hundred students attended at least one of three 
concerts: Pep Band, Country Sizzle, and Blue Mood. 48 heard the Pep 
Band concert. 36 heard the Country Sizzle concert. 60 heard the Blue 
Mood concert. 12 heard the Pep Band and Country Sizzle concerts.  
20 heard the Country Sizzle and Blue Mood concerts. 16 heard the Pep 
Band and Blue Mood concerts. How many attended all three concerts?

Question 5. A pair of dice features one die with three 0's and three 
1's on its faces and another with the numbers 1, 3, 5, 7, 9, and 11 on 
its faces.  This pair has the unusual property that only the sums 1, 
2, 3, . . . , 12 are possible and are equally likely to occur.  Design 
another such pair of six-faced dice that share this property.

Date: 04/13/97 at 15:51:59
From: Doctor Anthony
Subject: Re: More questions on probability

Question 4. Draw a Venn diagram as follows:  The universal set is a 
large square with the number 100 under the letter E (just outside the 
square).  Now draw three overlapping circles inside the square, and 
outside the first circle put the letter P and close under this the 
number 48. Outside the second circle put the letter C, and under this 
the number 36. Outside the third circle put the letter B and under the 
B put the number 60. Now where the three circles overlap put the 
letter x.  

We can now fill in the spaces with the number of students appropriate 
to that space. The overlap of P and C must total 12, but part of the 
overlap is already given as x. So in the remaining part put 12 - x.  
In a similar manner fill in the remaining part of the overlap of C 
and B with 20 - x, and the remaining part of the overlap of B and P 
with 16 - x. 

We can now fill in the empty space still in C, since it must total 36, 
and so the empty space will be x+4. Similarly the empty space in B 
will be x + 24.  We could find the empty space in P if necessary, but 
for one circle (I am going to use P for this purpose) it is sufficient 
to know that the total is 48.

We now add up all the space totals and put that total equal to 100.

Thus 

   48 + 4+x + 20-x + 24+x = 100 
   96 + x = 100        and so x = 4.

Therefore we have 4 students attending all three concerts.


Question 5. If you want the numbers 1 to 12 again, you could have one 
die with 2, 4, 6, 8, 10, 12   and the second with 0, 0, 0, -1, -1, -1.       

This might be cheating, but it gives the correct result.


-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 04/13/97 at 18:15:43
From: ac
Subject: More probability

Question 6.  Three vertices of a regular hexagon are randomly 
selected.  What is the probability that an isosceles triangle would be 
formed by connecting these three vertices?

Date: 04/14/97 at 07:40:38
From: Doctor Anthony
Subject: Re: More probability

Question 6. You can select the first vertex anywhere you like with 
probability 1. If we are to form an isosceles triangle we must select 
two more vertices which will form the triangle such that it is 
isosceles.  If you draw the figure it is easy to see that for any 
particular vertex there are four isosceles triangles of which it could 
be part, either as a base vertex or the vertex on the line of symmetry 
of the isosceles triangle.

For any one of these four triangles, the second vertex has a 2/5 
probability of being selected, and the third then has a 1/4 
probability of being selected.

   Total probability  = 4 x 2/5 x 1/4 = 2/5.


-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/