Translation of "ビット誤り確率" to English language:


  Dictionary Japanese-English

ビット誤り確率 - 翻訳 :

  Examples (External sources, not reviewed)

アルファが第一種の過誤の確率
And this is a slightly confusing notation, forgive me.
確率1 は確率40 よりも極端であり
The smallness of that probability is what we mean by extremity.
もし上の行を見ると 第一種の過誤の確率が分かり
So again each of these rows are sort of an alternate universe so if we, if we just
ユーロ圏の危機は確かに非常に誤解のビットと再出現をもたらした
Italians. This is precisely meant to avoid backlashes.
そして第二種の過誤の確率は 1 検出力 です
It's your probability of rejecting the null when in fact you should.
次に確率変数Xがあり確率は0 2です
What's the probability of the joint X, Y?
条件付き確率表によると50 の確率で曇りで 50 の確率で曇りません
In this case, there's only one such variable, Cloudy.
任意の確率変数Xがあり確率は0 2です
Question 1 In the first question, I'm going to ask you some very basic probability questions.
50 の確率 10 25 の確率 20
Then the value of the state for the action go up would be obtained as follows.
確率
Probability
確率?
Phil, the odds against
確率変数がある値に等しい確率 とか ある値より大きい(または小さい)確率 あるいは 確率変数が特定の性質を持つ確率
And it makes much more sense to talk about the probability or random variable equaling a value, or the probability that it is less than or greater than something or the probability that is has some property
成功確率 失敗確率です これが分散になります
And we know that our variance is essentially the probability of success times the probability of failure.
確率文法 ディープラーニング マルコフ確率場 他にもいろいろあります
It's kind of the big thing in application and machine learning.
求めた確率は 非正規化確率の1 αになります
Then I just normalize.
確率は
What are the odds?
8ビットだけのチャンネルでは 正確な8ビットの値は返ってきません
If you try to do it yourself you have to be aware of precision problems with your input data.
その確率を見積もる事が出来る つまり 研究者として第一種の過誤と第二種の過誤に陥るリスクが
What's nice is given the assumptions of the normal distribution and using normalized scores, we know the probability of winding up in one of those four boxes or we can estimate them.
PERFECT の確率は0にならず STORM の確率も0になりません
That is not the case for song.
別の確率を求めてみましょう スパムの確率とハムの確率です
Let's use the Laplacian smoother with K 1 to calculate the few interesting probabilities
成功確率
Probability of success
失敗確率
Probability of failure
確率が下がりましたね
Of course it has no clue that it is exactly over here.
つまり キューブを得る確率は
This represents the set of all the cubes.
予想どおりほとんどのセルの確率は0 03です 0 13の確率のセルもありますが このセルの確率は0 533です
Let's just check, and as predicted, almost all cells have a probability of 0.03.
なので 裏になる確率は 100 表の確率
And these are mutually exclusive events, you can't have both of them
どちらかが高いか低いか確率的誤差がなく 仮説検定は必要ありません
And comparing heights of the tallest building in two cities, that's a deterministic number.
正確に1を得る確率 掛ける 3 2を得る確率 3 3を得る確率かな 正確に1を得る確率 掛ける 3 2を得る確率 3 3を得る確率かな ですが 前回の動画を見ていれば
You might say OK, that's the probably of getting exactly 1 times the probability of getting 2 out of 3 plus the probability of getting 3 out of 3.
2つの確率変数XとYがあり それぞれの確率は0 2です
What's the probability of the complement?
事後確率を求めるため この出力の確率に事前確率を掛けます
We now apply Bayes rule.
コイン1を選ぶ確率がp0 表が出る確率がp1 1 p0でコイン2を選ぶ確率
And here is my answer. You can really read off the formula that I just gave you.
95 の確率で
If I pick a random T value, if I take a random T statistic
0.1 の確率で
There's going to be a 10 percent chance you get a pretty good item.
何が確率の...
Now let's have something a little bit more interesting.
確率ですと
Frack the odds.
同じ確率で
Equally possible,
最初に確率ノードがあります
Now this format begins to look very much like the game tree that we talked about in the previous unit.
確率を50 と見積りました
He estimated a probability that we will fail to survive the current century
つまり この確率は0です
They cannot happen at the same time.
雨の時に上機嫌である確率と雨である確率を掛けて 上機嫌である確率で割ります 1日目に雨である確率は このマルコフモデルから分かります
This is being answered using Bayes rule, so this is the probability of being happy given that it rains times the probability that it rains over the probability of being happy.
OLDの確率P OLD が事前確率となり クラスの数が基準となります OLDの中で Top の文字がある確率は P Top OLD そしてNEWの映画についての確率です
Use Laplacian smoothing with k 1 to compute the probability of a movie being old this is a prior probability, which is just based on class counts the probability of the word top as a title word in the class of old movies, and the probability that a new movie that we look at by new I mean a movie we've never seen before that is called top, the probability this movie that corresponds to the old movie class with the new movie class.
Aは0 9の確率でそのままAにとどまり 0 1の確率でBに遷移します そしてBは0 5の確率でBにとどまり 0 5の確率でAに遷移します
Here we're given a marchov chain between A and B, with the transition of A to itself is 0.9 with 0.1 chances transitions to B.
ある範囲の出現確率が得られる確率は 小さくなりません
Now, this statement is not true.
AでX3が成立する確率 AでX2が成立する確率 AでX1が成立する確率 Aが成立する確率です
If I keep expanding this, I get the following solution.
これらが終端値で 確率ノードには確率が等しい2通りの枝があります
One more quick game tree to evaluate.

 

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