0
Probability
The probability of a certain event is a number expressing the likelihood that a specific event will occur, expressed as the ratio of the number of actual occurrences to the number of possible occurrences. In mathematics, it is a measure of how often an event will happen and is the basis of statistics.
14,643 Questions
If you flip a coin 4 times how many possible outcomes are there?
There are 24 = 16 ordered outcomes, that is outcomes in which the order of the results is relevant. If not, there are 5 outcomes (0 heads, 1 head, 2 heads, 3 heads and 4 heads).
The mean of a binomial probability distribution can be determined by multiplying?
The mean of a binomial probability distribution can be determined by multiplying the sample size times the probability of success.
What is the probability of 6 in 30 numbers?
that is not even close to a question
6 occures in first 30 numbers for just 1 time so
probabability of 6 in 30 numbers= 1/30
95% CI for Proportion is p +/- E; where E = Z*sqrt(p*q/n). P=.6, q=.4, n=144, Z=1.96. Therefore E=1.96*sqrt(0.4*0.6/144) = 0.08. So, 95% CI = .6 +/- 0.08 or .52 to .68.
If you toss a quarter 3 times how many different orders can heads and tails be tossed?
There are 23 possible outcomes this is found by there being 2 possibilities ( head or tails) that is done three times, that is 2X2X2= 8 possible combinations for the outcome. This is if order matters, that is h h t is different to t h h
What is the difference between standard error of mean and standard deviation of means?
Standard error of the mean (SEM) and standard deviation of the mean is the same thing. However, standard deviation is not the same as the SEM. To obtain SEM from the standard deviation, divide the standard deviation by the square root of the sample size.
at least 39 blue marbles
If you roll a die twice what is the probability of getting a summation greater than 7?
There are 36 permutations of rolling a die twice; 11, 12, 13, 14, 15, 16, 21, 22, 23, 24, 25, 26, 31, 32, 33, 34, 35, 36, 41, 42, 43, 44, 45, 46, 51, 52, 53, 54, 55, 56, 61, 62, 63, 64, 65, and 66. (The first digit is the first die, and the second digit is the second die.)
Of these, 15 have a sum of greater than 7; 26, 35, 36, 44, 45, 46, 53, 54, 55, 56, 62, 63, 64, 65, and 66.
The probability, therefore, of rolling a sum greater than 7 with one die rolled twice (or two dice rolled once) is 15 in 36, or 5 in 12, or 0.4166666... .
To determine the probability of picking 3 cards of one suit and 1 card of another in a standard 52 card deck, consider each card one at a time.
The probability of picking a card in any suit is 52 in 52, or 1.
Since there are now only 12 cards in the first suit, the probability of picking a card in the same suit is 12 in 51, or 4 in 17, or 0.2353.
Since there are now only 11 cards in the first suit, the probability of picking a card in the same suit is 11 in 50, or 0.22.
Since there are still 39 cards in the remaining three suits, the probability of picking a card in a suit different than the first is 39 in 49, or 0.7959.
The probability of picking 3 cards of one suit and 1 card of another in a standard 52 card deck is, therefore, the product of the probabilities of each card, or (52 in 52) (12 in 51) (11 in 50) (39 in 49), or 267696 in 6497400, or 0.0412, or about 1 in 25.
.9^27, or approximately .058 = 5.8%
For me I think it is conceptually easier to think about the probability that the number will contain the digit seven (and the probability that it does not contain the digit 7 is simply one minus the probability that it does).
P(number will contain 7) = P(number is in the seven hundreds) + P(number is not in seven hundreds)*[P(number is in the X hundred seventies)+P(number is not in the X hundred seventies)*P(number ends in seven)]
So essentially I am considering all of the numbers in the range that start with seven (i.e., are in the seven hundreds), then all of the numbers in the range that aren't in the seven hundreds but have a 7 in the tens place (i.e., the 170s, 270s, etc., and finally all the numbers that don't have a 7 in the hundred or tens place, but that end in 7).
Plugging the numbers into my formula above, I get (100/900)+(800/900)*((10/100)+(90/100)(1/10)) = 7/25 is probability that the number does contain a 7, and 1-(7/25)=18/25 is probability that it does not.
How many permutations are possible using the 6 letters in the word Alaska?
120 if A is treated the same as a.
360 if A is considered different to a.
(13/52)*(12/51)*(11/50)*(10/49)*(9/48)*(8/47)=33/391510, which is approximately .0000842 .
If 15 strangers are all in a room what is the probability of them all having the same birthday?
To determine the probability of 15 random people all having the same birthday, consider each person one at a time. (This is for the non leap-year case.)
The probability of any person having any birthday is 365 in 365, or 1.
The probability of any other person having that same birthday is 1 in 365, or 0.00274.
The probability, then, of 15 random people having the same birthday is the product of these probabilities, or 0.0027414 times 1, or 1.34x10-36.
Note: This answer assumes also that the distribution of birthdays for a large group of people in uniformly random over the 365 days of the year. That is probably not actually true. There are several non-random points of conception, some of which are spring, Valentine's day, and Christmas, depending of culture and religion. That makes the point of birth, nine months later, also be non-uniform, so that can skew the results.
False. A normalized distribution curve (do not confuse normalized with normal), by definition, has an area under the curve of exactly 1. That is because the probability of all possible events is also always exactly 1. The shape of the curve does not matter.
2/50 x 2/50= 4/2500= 1/625
-------------------------
If the first ticket won't be replaced after it has been drawn, then your chances are 2/50*1/49=2/2450=1/1225.
orgy
first you have to keep in mind that probability is # of favorable outcomes/ # of possible outcomes. we already know (given that no two students are the same height) that there is only 1 possible way to order the students from shortest to tallest therefore, we have established that we have 1/# of possible outcomes. to find the number of possible outcomes you take 8 factorial. 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40320. now you have the # of favorable over # of possible = 1/40320 = .0000248015873 = 2.48015873 E-5.
What are some examples of decimals in everyday life?
hey dude somewhat right answer but read
like when you buy stuff the prices are usually decimals..
. like $3.99
your body temperature at a normal temp is 98.6 which is a decimal...
