course Mth 152 I apologize for not using the form to get your attention on the broken links. If I run into that problem in the future I will do so. Thanks for the help in getting access to the assignments.
.............................................
Given Solution: The probability of obtaining a 3 on a single role is 1/6 (one of the six possible outcomes is a 3). Since the two rolls are independent, it follows that if two dice are rolled the probability of obtaining two 3's is 1/6 * 1/6 = 1/36. Self-critique: I understood this pretty well. ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q002. What is the probability that on three rolls of a fair die, we obtain exactly two 5's? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Getting two 5’s on three rolls of a die is 1/6*1/6 and for the third(second or first works as well) roll, we have a 5/6 chance of not getting a 5. Therefore, we have 1/6*1/6*5/6 or 5/216 chance of getting exactly two 5’s on three rolls. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: On three rolls of a fair die, the two 5's can occur on the first and second, the first and third or the second and third rolls. That is, of the three available positions the two positions in which the 5's occur can occur in C(3,2) = 3 different ways. Since the probability of a 5 on any roll is 1/6 and the probability of not getting a 5 on a roll is 5/6. Any one of the three ways of getting two 5's and one non-5 is therefore (1/6) * (1/6) * (5/6 ) = 5/216. Since each of the three ways to get the desired outcome occurs with probability 5/216, it follows that Probability of exactly two 3's on three rolls = 3 * 5/216 = 15/216 = 5/72. STUDENT QUESTION: I’m confused as to why you multiply the 5/216 by 3? Doesn’t the 5/216 give you the correct answer? 1/6 *1/6 *5/6 shows the probability of all three dice. INSTRUCTOR RESPONSE: 1/6 *1/6 *5/6 is the probability of getting 5 on the first die, 5 on the second and something else on the third. However you can also get 5 on the first, something else on the second, and 5 on the third. The probability of this outcome could be written 1/6 * 5/6 * 1/6, showing the order of the three events. Or you could get something else on the first and 5 on each of the last two. The probability of this outcome could be written 5/6 * 1/6 * 1/6, again showing the order of the three events. When multiplied out, the probability of any of these three events is 1/6 * 1/6 * 5/6. The three events are mutually exclusive, so the probability that one of the three events will occur is 3 * 1/6 * 1/6 * 5/6. Self-critique: After reading the response above, I understand why you multiplied this by 3 because of the ability to get two of the correct choices in any order. ------------------------------------------------ Self-critique rating:2 ********************************************* Question: `q003. What is the probability that on six rolls of a fair die, we obtain exactly two 5's? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To obtain two 5’s we have a 1/6 *1/6 chance in addition to multiplying this by 5/6*5/6*5/6*5/6 due to the chances of not getting a 5. So for this we have 6*1/6*1/6*5/6*5/6*5/6*5/6 like the previous problem. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: In order to get exactly two 5's on six rolls of the fair die, we must get two 5's and four results that are not 5. The probability of getting a 5 on any roll is 1/6, and the probability of getting a result other than 5 is 5/6. Therefore given any two positions out of the six the probability of obtaining 5's in two of the positions and non-5's in the remaining four positions is by the Fundamental Counting Principle Probability of 5's in exactly two of the six positions = (1/6) * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) = (1/6)^2 * ( 5/6)^4. There are C(6,2) ways in which the positions of the two 5's can be selected from the six available positions. Thus we have Probability of exactly two 5's on six flips = C(6,2) * (1/6)^2 * (5/6)^4. STUDENT COMMENT: So, because our outcome (two 5’s) can only occur 2 of the 6 times, in any order, we have to also multiply by C(6,2). This is the same for the previous question, which was multiplied by 3 (or C(3,2) ) This concept makes a lot more sense now. INSTRUCTOR RESPONSE: Right. The two 5's can occur in any two of the 6 rolls, and there are C(6, 2) ways of selecting which two. Self-critique: If I am reading this response correctly, I seem to have the correct idea in solving the problem above?
.............................................
Given Solution: By analogy with the preceding problem, we see that to get r 5's on n rolls we must get 5 the total of r times and non-5 a total of (n-r) times. Since probability of getting a 5 is p, the probability of getting 5 a total of r times is represented by p^r. Since the probability of getting a non-5 is q, then the probability of getting a non-5 a total of (n-r) times is represented by q^(n-r). There are C(n, r) ways to place fives in r of n positions, so the probability of getting 5 fives and n non-fives is C(n, r) * p^r * q^(n-r). Self-critique: I understood this problem pretty well but was still slightly confused about the q^(n-r) part.
.............................................
Given Solution: If we roll a single die, we either get 5 or we don't. The two events are mutually exclusive -- they can both happen on the same roll. They also cover all possibilities. The sum of the probabilities is therefore 1. So we conclude that p + q = 1, and from this it follows immediately that q = 1 - p. Substituting 1 - p for q in the expression C(n, r) * p^r * q^(n-r) we obtain Probability of r fives on n rolls = C(n, r) * p^r * (1-p) ^ (n-r). Ok I have this wrong in the problem and am a bit confused. Why is the problem *p^r*(1-p)….?