#$&* course MTH 152 3/6 11 009. Binomial Probabilities*********************************************
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Given Solution: The probability of obtaining a 3 on a single role is 1/6 (one of the six possible outcomes is a 3). Since the two rolls are independent, it follows that if two dice are rolled the probability of obtaining two 3's is 1/6 * 1/6 = 1/36. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q002. What is the probability that on three rolls of a fair die, we obtain exactly two 5's? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 6 possible outcomes for the first roll of one fair die, 36 possible outcomes for the second roll of one fair die, and 216 possible outcomes for the third roll of one fair die. There are three possible ways in which we can get exactly two 5's: [5,5,-], [5,-,5], and [-,5,5]. 5 out of 216, multiplied by the number of rolls since order does not matter, is 3*5/216, or 15/216. Therefore, the probability of obtaining exactly two 5's on three rolls of a fair die is 15/216, or 5/72. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q003. What is the probability that on six rolls of a fair die, we obtain exactly two 5's? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 6 possible outcomes for the first roll of one fair die, 36 for the second roll of one fair die, 216 for the third roll of one fair die, 1,296 possible outcomes for the fourth roll of one fair die, 7,776 possible outcomes for the fifth roll of one fair die, and 46,656 possible outcomes for the sixth roll of one fair die. There are 15 possible ways in which we can get exactly two 5's: [5,5,-,-,-,-], [5,-,5,-,-,-], [5,-,-,5,-,-], [5,-,-,-,5,-], [5,-,-,-,-,5], [-,5,5,-,-,-], [-,5,-,5,-,-], [-,5,-,-,5,-], [-,5,-,-,-,5], [-,-,5,5,-,-], [-,-,5,-,5,-], [-,-,5,-,-,5], [-,-,-,5,5,-], [-,-,-,5,-,5], and [-,-,-,-,5,5]. 15 out of 46,656, multiplied by the number of rolls since order does not matter, is 6*15/46,656, or 90/46,656. Therefore, the probability of obtaining exactly two 5's on six rolls of a fair die is 90/46,656 or 5/2592. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT COMMENT: If I am reading this response correctly, I seem to have the correct idea in solving the problem above? INSTRUCTOR RESPONSE: That is the right idea, but (1/6)^2 * (5/6)^4 would be the probability that you get the two 5's on two specified rolls (e.g., the probability that you get the 5's on rolls 2 and 4). You would also have a probability of (1/6)^2 * (5/6)^4 of getting the 5's on rolls 1 and 4, as well as the probability of getting the 5's on rolls 2 and 5, etc.. You could get the 5's on rolls 1 and 2, or on rolls 1 and 3, or on rolls 1 and 4, or on rolls 1 and 5, or on rolls 1 and 6, or on rolls 2 and 3, or on rolls 2 and 4, or on rolls 2 and 5, or on rolls 2 and 6, or on rolls 3 and 4, or on rolls 3 and 5, or on rolls 3 and 6, or on rolls 4 and 5, or on rolls 4 and 6, or on rolls 5 and 6. That is C(6, 2) = 15 ways to get the two 5's, each with probability (1/6)^2 * (5/6)^4. So the probability is C(6, 2) * (1/6)^2 * (5/6)^5. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q004. If we let p stand for the probability of getting a 5 on a roll of a die and q for the probability of not getting a 5 on a roll, then how would we expressed a probability of getting exactly r 5's on n rolls? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find r number of 5's on n number of rolls, we must calculate how many times r number of 5's can be obtained on n number of rolls. We can obtain 5 the total of r times and not-5 a total of n-r times. If p stands for the probability of r and q stands for the probability of n-r times, and C(n,r) stands for the number of ways to place fives in r of n positions, we can construct an equation of C(n,r)*p^r*q^(n-r). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). STUDENT COMMENT: I understood this problem pretty well but was still slightly confused about the q^(n-r) part. INSTRUCTOR RESPONSE On the previous example, n was 6 (for the 6 rolls) and r was 2 (the number of 'successful' outcomes, regarding a 5 as a success). The probability of a success was 1/6. Our expression was C(6, 5) * (1/6)^2 * (5/6)^4. Using the symbols n, r and p we would write this as C(n, r) * p^r * (5/6)^4. What about the 5/6 and the 4? 4 is the number of 'failures'. There were 6 rolls with 2 'successes' and therefore 6 - 2 = 4 'failures'. It should be clear that if there are n trials and r 'successes' there are (n - r) 'failures'. 5/6 in this example is the probability of 'not getting a 5', i.e., the probability of a 'failure'. The probability of a success and the probability of a failure add up to 1 (there is a probability of 1, or 100%, that the trial will either succeed or fail). If we let q stand for the probability of a failure, we get p + q = 1, so that q = 1 - p. So our expression C(n, r) * p^r * (5/6)^4 generalizes to C(n, r) * p^r * q^(n-r), where q = 1 - p. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q005. Explain why, if p is the probability of getting a 5 on a single roll of a die, it follows that the probability q of not getting a 5 is q = 1-p. How would we therefore express the formula C(n, r) * p^r * q^(n-r) only in terms of p? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we are rolling a die only a single time, we can either obtain a 5 or obtain something other than a 5. That means that the possibility of rolling any number, 5 or not, is only 1. If the probability of obtaining a 5 is represented by p, then we can subtract that probability by the total number of possibilities overall. Therefore, the probability of not getting 5 is represented by q=1-p. This means that we can substitute 1-p for q in the equation we constructed in the previous problem, giving us an equation of C(n,r)*p^r*(1-p)^(n-r). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we roll a single die, we either get 5 or we don't. The two events are mutually exclusive -- they cannot 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). STUDENT COMMENT: I have this wrong in the problem and am a bit confused. Why is the problem *p^r*(1-p)….? INSTRUCTOR RESPONSE: Recall that q = 1 - p. So C(n, r) * p^r * q^(n-r) could be written as C(n, r) * p^r * (1-p)^(n-r) Self-critique: OK ------------------------------------------------ Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!