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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 2 times how many possible outcomes are there?
If you flip a coin 2 times, there are 4 possible outcomes; HH, HT, TH, TT.
P(B) = 0.12 = 12%
There are 25 pair of socks: 10 Black, 12 White, 3 Brown.
P(B) = 3/25 = 0.12 = 12.0%
If you sit on a moving bus and toss a coin straight up into the air where will it land?
The coin is traveling at the same velocity as the tosser, so it will land the same as if you were standing still.
A single card is drawn from a deck Find the probability of selecting a 10 or a diamond?
for the diamonds - you have a 1in4 chance (25%) 52 cards in the deck, 13 are diamonds or 13:52=1:4 for the 10s - you have 1in13 chance (7.7%) 52 cards in the deck, 4 are 10s or 4:52=1:13
What are the chances of rolling double 6 3 times in a row?
The chance of rolling a double 6 is 1/6x1/6
The chance of getting the 3 times in a row is
(1/6x1/6)3 or (1/36)3 which is 1/46,656, a very small number!
0.97. Just take it away from 1
If you flip a coin 8 times how many possible outcomes are there?
The following notes answer the above question completely and neatly. It's better in maths to try to keep it simple.
Looking at the different possible outcomes; let H = Head and T = Tail. Then the different possible outcomes are (8H +0T) meaning 8 Heads and Zero Tails, or (7H + 1T) or (6H + 2T) or (5H + 3T) or (4H + 4T) or (3H + 5T) or (2H + 6T) or (1H + 7T) or (0H + 8T) giving 9 different outcomes (possibilities).
The 9 different outcomes are not equally probable, (8H + 0T) and (0H + 8T) are the least likely while (4H + 4T) will be the most likely. However, totaling all possible outcomes there are 256.
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There are 256 possible "histories" of the whole session, corresponding to
the number of possible values of one binary 'byte', where each flip fills in
one bit of the byte.
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Another Answer:
As is usual in science, the answer depends on how you look at the problem, and on your definition of the terminology that you use. Nine different outcomes, as described above, is correct when you decide to consider only distinguishable or, more, correctly, combinatorial, outcomes.
However, in statistics and probability, we don't usually deal with combinations, we deal with permutations, because looking at permutations allows you to see the indistinguishable outcomes, outcomes which do, in fact, affect the probability of the result.
In this case, throwing a coin 8 times (or with identical results, throwing 8 coins one time), there are multiple permutations that yield the same combination. For instance, there are eight permutations of one coin being heads, there are 28 permutations of two coins being heads, and so on and so forth, leading up to a grand total of 256 different permutations of 8 coins, yielding "only 9 outcomes".
So, in summary, there are 9 distinguishable "outcomes", or combinations, while there are 256 indistinguishable "outcomes", or permutations.
(To clarify "indistinguishable", I should say that you could certainly mark each coin with a number, say from 1 to 8, and then you would be able to distinguish the outcomes or heads or tails as a function of which coin was thrown. Similarly, throwing only one coin at a time, you could write down the outcomes as they occur, and they would certainly be "distinguishable". In this context, "indistinguishable" simply means that, in the absence of some tracking system, a throw, for instance, of HTTTTTTT, or THTTTTTT, would not be different - they would both be counted as (1H + 7T), using the notation of the earlier answer.)
Now, to nail this down for completeness sake, because the answer does matter, even though the question did not ask it, take a look at probability...
Probability is simply the number of permutations of a desired result divided by the number of permutations of all results. Since there is one permutation of no heads (0H + 8T), the probability of no heads is 1 in 256, or about 0.00391. On the other hand, since there are 8 permutations of one head (1H + 7T), the probability of one head is 8 in 256, or about 0.0313. Going one more step, the probability of two heads is 28 in 256, or about 0.109.
Understanding the distinction between combinations and permutations, and distinguishable versus indistinguishable, is key to understanding probability.
What is the middle number between 29 and 35?
The number between 29 and 35 would be 32. This is 3 above 29 and 3 below 35.
First: it's not "twirl a pen", it's pen spinning. and there are many different tricks.
Second: join upsb.info it's a pen spinning forum
Third: you could also search for ktrinh93 's videos for tutorials
Forth: if you're gonna get a modded pen, please dont get a buster, beginners should not spin busters, you'll improve too fast and when you spin an mx or bictory later, you'll fail (usually)
GL fellow spinner :D
There are 26 total marbles. We will call x the number of blue marbles. So there are x+2 red marbles. If you add them together you get x + x + 2....and all that equals 26 (subtract 2 from each side and add the two x's together) 26 - 2 = x + x 24 = 2x then divide each side by two and you get 24/2 = 2x/2 simplified is 12 = x So x is 12 (blue marbles) and x + 2 is 14 is (the amount of red marbles). 12 + 14 = 26!
What is the probability of randomly selecting a heart from a standard deck of 52 card?
There are 13 hearts in the standard deck, A-10,J,Q,K. So your probability is 13/52 which is 1/4 or 25% chance
Let me denote -A as the event that A does not happen. So we want Pr[-(A and B)]
Now, the event that neither A nor B occurs is the opposite of either A occurring, or B occurring or both occurring.
So Pr[-(A and B)] = 1 - Pr(A or B)= 1 - [Pr(A) + Pr(B) - Pr(A and B)] (since A+B is double counted)= 1 - (0.5 + 0.7 - 0.4)= 1 - 0.8= 0.2
This is a Binomial Probability Distribution; n=4, p=0.6. The probability of at least 1 passed is equal to the probability of 1-none passed; so x=0. The probability of x=0 (with n=4, p=0.6) is 0.0256. So, the probability of at least 1 passed is 1-0.0256 or 0.9744.
What is the probability that a 70 percent free throw shooter would make 1 of 2 shots?
The probability the shooter makes both shots is .7 * .7 = .49, and the probability of making neither is .3 * .3 = .09. So the probability of making exactly 1 out of 2 is 1 - .49 - .9 = .42, or 42 percent.
What is the definition for an equally likely outcome in the probability of an event?
http://www.emathzone.com/tutorials/basic-statistics/definition-of-probability.html
If mean equals 18.6 standard deviation equals 4 what is the z score?
The z score, for a value y, is (y - 18.6)/4
Two six sided dice are rolled what is the probabiliy for the sum to be greater than 4?
There are 36 equally-likely possibilities for a 2-dice roll (six possibilities for each die, multiplied together). Let's consider the probability of the sum being less than or equal to 4. There is only one way to roll a 2 (1+1), there are two ways to roll a 3 (1+2, 2+1), and three ways to roll a 4 (1+3, 2+2, 3+1). So there are 6 ways to roll at most a 4, meaning that there must be 36-6=30 ways to roll more than a 4. The probabiliity of rolling a sum greater than 4 is therefore 30/36, or 5/6.
What is the probability of drawing a king and a diamond from a deck of playing cards?
The probability of drawing a king is 4:52
The probability of drawing a diamond is 13:52 (or 1:4)
The probability of drawing a king (0.07692...) then replacing that king into the deck then drawing a diamond is 0.019230769.
If you leave the king out, the probability will be slightly greater (4/52) * (13/51)
Unless the king you left out of the deck was a king of diamonds, in which case, the probability would be (4/52) * (12/51)
How do you find the area to the right of the z-score?
Charts typically show and list the area to the left of the Z-Score value. To find the area to the right, just subtract the Z-Score value from 1; e.g. if the Z-Score value is .75 then take 1-.75 = .25.
10*9*8=720
The sample space is 23 or 8; which can be listed out as: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. There are 2 of the 8 that have exactly 2 heads; so the probability of exactly two coins landing on heads is 2/8 or 1/4.
