http://www.pimp-your-poker.com/poker-hands-scores/
Probability problems?
I have some problems I'm stuck on a probability. I know what the answers are problems because of practice test, but I'm always different answers, and I am confused how to solve them. A coin is flipped 4 times. Find the probability at least 1 head. -> Answer 0.9375. I think that should be asnwer 1 / 8 A poker hand is dealt five cards 52 in any order from an ordinary Deck. Find the probability that a poker hand consists of 3 hearts and 2 clubs. -> Answer 143/16660 A Student ID consists of 7 digits. How many students ID's are there when retry is not permissible under the digits, and the student ID must be 0 begin with? -> 480th reply 60 I thought it was 151 200 (1x10x9x8x7x6x5). A multiple choice test consists of 20 questions with 5 choices for each question. If a student guesses on each question, what is the probability of a value from zero to the test: -> Answer 0.0115 Any help is greatly appreciated. Thanks:)
1) the difference in your answer and what is the question we must ask that the Question P's claim (X ≥ 1) where, as your answer for P (X = 1) Let X coin, the number of heads in n tosses of a fair view. X has the binomial distribution with n = 4 trials and success probability p = 0.5 In general, if X is the binomial distribution with n trials and a success probability of p then P [X = X] = n! / (X! (nx)!) * P ^ x * (1-p) ^ (nx) for values of x = 0, 1, 2, …, n P [X = x] = 0 for every other value of x. The probability of n Objects and then sum total of x success and n -. X or failure, in other words, the binomial is the function, the mass is derived by dividing the number of combination of objects x of n independent and identically distributed Bernoulli trials. X ~ Binomial (n, p) the mean of the binomial distribution is n * p = 2 the variance of the binomial distribution is n * p * (1 – p) = 1, the standard deviation the square root of the variance = √ is (n * p * (1 – p)) = 1 The probability mass function, PMF, f (X) = P (X = x) P (X = 0) = 0.0625 P (X = 1) = 0.25 P (X = 2) = 0.375 P (X = 3) = 0.25 P (X = 4 ) = 0.0625, the cumulative distribution function, CDF, F (X) = P (X ≤ x): x Σ P (X = t) = t = 0 P (X ≤ 0) = 0.0625 P (X ≤ 1) = 0.3125 P (X ≤ 2) = 0.6875 P (X ≤ 3) = 0.9375 P (X ≤ 4) = 1 1 – F (X): n Σ P (X = t) = t = x P (X ≥ 0) = 1 P (X ≥ 1) = 0.9375 P (X ≥ 2) = 0.6875 P (X ≥ 3) = 0.3125 P (X ≥ 4) = 0.0625 == – == – == – == – == – == – == 2) If you have n objects and chose r of them, the number of combinations is this: n! / (R! (nr)!) NCr this can be written as there are 52 C 5 = 2.59896 million possible five-card hands are (13 C 3) * (13 C 2) = 22 308 hands == From three hearts and two clubs 22308 / 2598960 = 143/16660 = 0.008583433 – is == 3) no repeat of the first spot as a zero – == – == – == – == – == and the second digit can be 1, 2, 3, …, 9, but not zero 1 * 9 * 8 * 7 * 6 * 5 * 4 = 60 480 == – == – == – == – == – == – == 4) to test received a score of zero on the The student must answer the question 20 wrong. there is a 4 / 5 probability of answering the questions incorrectly. wrong answer all 20, you have: (4 / 5) ^ 20 = 0.01152922 in a general remark, give X the number of correct answers. X ~ Binomail (n = 20, p = 5.1) P (X = 0) = 0.01152922 P (X = 1) = 0.05764608 P (X = 2) = 0.1369094 P (X = 3) = 0.2053641 P (X = 4) = 0.2181994 P (X = 5) = 0.1745595 P (X = 6) = 0.1090997 P (X = 7) = 0.05454985 P (X = 
= 0.02216088 P (X = 9) = 0.007386959 P (X = 10) = 0.002031414 P (X = 11) = 0.0004616849 P (X = 12) = 8.656592e-05 P ( X = 13) = 1.331783e-05 P (X = 14) = 1.664729e-06 P (X = 15) = 1.664729e-07 P (X = 16) = 1.300570e -08 P (X = 17) = 7.65041e-10 P (X = 18) = 3.187671e-11 P (X = 19) = 8.388608e-13 P (X = 20) = 1.048576e-14
The probability of a great poker hand
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