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/
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