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Now consider a guessing game where my friend picks a number between 1 and 100. And I need to guess what it is, asking only simple yes or no questions of the form, is your number bigger than x? My best strategy is going to be to eliminate half of the remaining possibilities each time. Dear Stats Person, Here is my question. If a person were to guess a number from 1-10, the chance of guessing a 4 would be 0.1. If 5 people were a. . . BINOMDIST, Binomial Probability, Binomial pmf # 419 :: 9/28/08: Suppose two events are independent. One event has probability of 0.25, while the other has probability of 0.59.. . . Multiplication ...

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Aug 01, 2009 · The Vardon Exploration Company is getting ready to leave for South America to explore for oil. One piece of equipment requires 10 batteries that must operate for more than 2 hours. The batteries being used have a 15% chance of failing within 2 hours. The exploration leader plans to take 15 batteries. Assuming that the conditions of the binomial distribution apply, the probability that the ...
Probability is defined as a proportion, and it always takes values between 0 and 1 (inclusively). It may also be displayed as a percentage between 0% and 100%. Probability can be illustrated by rolling a die many times. Consider the event “roll a 1”. The pattern evident from parts (a) and (b) is that if K + 1 dogs are boarded together, one a carrier and K healthy dogs, then the probability that at least one of the healthy dogs will develop kennel cough is P (X ≥ 1) = 1 − (0.992) K, where X is the binomial random variable that counts the number of healthy dogs that develop the condition.

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The minor difference is that odds is stated as 1 in some number; and odds against when that number is less than 1; and probability is generally stated as a decimal or percentage. In history, all probabilities have become 0 or 1; until we know whether the event happened or not, its chance of happening, or probability will be a number between 0 ...
The chance that any two specific people will choose the same number is 1/100. How many opportunities were there for such an event to occur? There are 20 people, making 20$\times$19/2 = 190 pairs of people. What is the expected number of events? This is 190/100= 1.9. 4. 2. 0. or 7, 7, 10, 13 wherein the difference between the 2nd & the 3rd or the 3rd & the 4th is more than 1, come under lower correlation type The formats observed as 1, 1, 2, 3, or 2, 2, 3, 4, or 4, 4, 3, 2 or 7, 7, 8, 9 wherein the difference between the said consecutive numbers is 1, come under the normal correlation.

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For this problem, students can apply the probability formula introduced in the Opening: P(A) = a n where P(A) = the probability of rolling a 3, a = 1 (the number of ways rolling a 3 can occur), n = 6 (the total possible equally-likely outcomes when rolling a number cube), and P(rolling a 3) = 1 6.
Row A: Invalid. Sum is greater than 1. Row B: Valid. Probabilities are between 0 and 1, and they sum to 1. In this class, every student gets a C. Row C: Invalid. Sum is less than 1. Row D: Invalid. There is a negative probability. Row E: Valid. Probabilities are between 0 and 1, and they sum to 1. Row F: Invalid. There is a negative probability. 2.5.1 Number of subsets; 2.6 ... is equally likely to take any value between 0 and 1. Formally, the probability density ... find the probability that Frank will guess ...

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Aug 31, 2005 · Reply to Christopher Valk's Post: that is incorrect and i am disappointed that i have to pay for this because i need this answer for my studies and now i need to re ask it. the chances of my guessing ten sequential coin flips in a row correctly are 1 in ten thousand. you dont just scale it down like a fraction. or the odds of me guessing one coin flip would be 1 in 1000 according to this ...
find the probability of guessing correctly at least 6 of the 10 answers on a true or false examination. the correct answer is 193/512 please explain and show all working. thanks very much! Sep 13, 2008 · And for the correct guessing in the second guess, we only have 9 options left for X, since we will not choose the same number as we have guessed in te first guessing. So, it'll be 1/9. So, the probability for this case is 9/10 * 1/9 = 1/10. case 3: Correct guessing in the second guessing

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Then any string of words all of which appear in the corpus will be assigned a probability of (1/ 47,885 )N, where N is the number of words in the string, assuming a sample space of sentences all of length N. A sentence of 6 words would be assigned a probability of (1/ 47,885)6, which just so happens to be about (2.1 * 10-5)6, or 86 * 10-30.
You are to participate in an exam for which you had no chance to study, and for that reason cannot do anything but guess for each question (all questions being of the multiple choice type, so the chance of guessing the correct answer for each question is 1/d, d being the number of options per question; so in case of a 4-choice question, your ... The probability of a packet reception from S to R is 1-p and the probability of an ACK reaching S given that R sent an ACK is 1-q. The sender moves from sequence number k to k + 1 if the packet reaches and the ACK arrives.

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(iv) From numbers 1 to 100, there are 15 numbers which are greater than 85 i.e. {86, 87, …., 98, 99, 100} So, favorable number of events = n(E) = 15. Hence, probability of selecting a card with a number greater than 85 = n(E)/ n(S) = 15/100 = 3/20 (v) From numbers 1 to 100, there are 47 numbers which are less than 48 i.e. {1, 2, ……….., 46, 47} So, favorable number of events = n(E) = 47
3) A maximum period size for patterns to report. Period size is the program’s best guess at the pattern size of the tandem repeat. The program will find all repeats with period size between 1 and 2000, but the output table can be limited to some other range. 4) A minimum alignment score to report repeat. Some mathematical uses include the probability of rolling a six on a die, the probability of tossing a coin and getting “heads,” and the probability of 1-2-3 coming up as the daily lottery number. Problem A2 Statistical uses of probability include the probability that the estimate of a mean is accurate (this is known as a confidence interval).

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Person A chooses an integer between 1 and 100 at random, then B can guess that number in (at most) 7 attempts, i.e. $\log_2(100)+1=7$. What if now A chooses an integer from a distribution that is known to B (B knows the probability of each number being selected but does not know the number).
• 0 ≤ P(A) ≤ 1, the probability of an event must be a number between 0 and 1, inclusive • P(S) = 1, the probability of the entire sample space must be equal to 1 • If an event must happen, P(A) = 1 • If an event is impossible, P(A) = 0

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Feb 01, 2017 · BUS 308-The probability of finding 100 defective products - Subject Statistics - 00474733 ... our best guess. based upon counting the number of times an event occurs ...
The number of ways to choose 5 out of the 6 winning numbers is given by 6 C 5 = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by 42 C 1 = 42. Thus the number of favorable outcomes is then given by the Basic Counting Rule: 6 C 5 × 42 C 1 = 6 × 42 = 252. So the probability of winning the second prize is