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AQA · GCSE · Mathematics נושא 5 3:32 קריאת קול באנגלית · תרגום אנגלי + סינית שרוף בתוך הסרטון
פרקים
תרגום טקסט
Four families carry probability.
概率有四类要点。
Every probability lies from 0 to 1, and an exhaustive set of outcomes sums to 1, so the complement is 1 minus P.
每个概率都在 0 到 1 之间，一组穷尽的结果概率之和为 1，所以对立事件是 1 − P。
From experiments, relative frequency estimates P, and n times p is the expected count.
由试验得到的相对频率估计 P，n × p 是期望次数。
Independent events multiply; dependent ones change the denominator.
独立事件相乘；不独立的事件要改变分母。
And a condition restricts the denominator to the given group.
条件把分母限制在给定的那一组。
Toss a fair coin.
抛一枚均匀硬币。
6 heads in 10 estimates 0.6, far from a half.
10 次中 6 次正面，估计为 0.6，离一半很远。
502 in 1000 estimates 0.502, much closer.
1000 次中 502 次，估计为 0.502，接近得多。
Larger samples are more stable, but never exact.
样本越大越稳定，但永远不精确。
So when four students' results disagree, the best estimate comes from the one with the most flips.
所以四名学生的结果不同时，最好的估计来自抛得最多的那位。
The chance of rain on a June day is 0.38.
六月某天下雨的概率是 0.38。
How many of 150 days do we expect to be dry?
150 天中预计有多少天不下雨？
Dry is the complement: 1 minus 0.38 is 0.62.
不下雨是对立事件：1 − 0.38 = 0.62。
Multiply by the number of days: 0.62 times 150 is 93.
乘以天数：0.62 × 150 = 93。
That is an average over many Junes; any single year can differ.
这是多年平均；任何一年都可能不同。
A spinner lands red, blue or green, and nothing else.
一个转盘只会停在红、蓝、绿上。
Red is 0.3 and blue is 0.45.
红 0.3，蓝 0.45。
Lay them along the scale from 0 to 1: the pieces must fill it exactly, so green must be 1 minus 0.75, which is 0.25.
把它们沿 0 到 1 的标度排开：它们必须恰好填满，所以绿色是 1 − 0.75 = 0.25。
A list that adds to 1.1 is impossible.
加起来是 1.1 的概率表是不可能的。
Two dice give 36 ordered pairs, all equally likely.
两个骰子有 36 个有序结果，都等可能。
Write each total in the grid.
在表格里写出每个和。
Exactly six sum to 7, on the diagonal, so the probability is 6 out of 36, a sixth.
恰好六个和为 7，在对角线上，所以概率是 6/36 = 1/6。
At least 11 is three cells: 1 in 12.
至少 11 有三格：1/12。
Order matters: 3 then 4 differs from 4 then 3.
顺序有关：先 3 后 4 与先 4 后 3 不同。
A class of 30: 18 cycle, 12 swim and 7 do both.
一个 30 人的班：18 人骑车，12 人游泳，7 人两样都做。
Put 7 do both in the overlap first.
先把两样都做的 7 人放进重叠部分。
Then cycling only 11 and swimming only 5.
然后只骑车 11 人，只游泳 5 人。
At least one is 23, so 7 do neither.
至少一样是 23 人，所以两样都不做的有 7 人。
Both means the overlap; either means the whole of the two circles.
“都”指重叠部分；“任一”指两个圆的全部。
Justify independence from context before multiplying bare products.
先用情境说明独立性，再做纯乘积。
Picture 1000 people.
设想 1000 人。
20 percent have a condition: 200 do, 800 do not.
20% 有某种病：200 人有，800 人没有。
The test is positive for 90 percent of the 200 and 10 percent of the 800: 180 plus 80, so 260 positives.
检测对 200 人中的 90% 呈阳性，对 800 人中的 10% 呈阳性：180 + 80，共 260 个阳性。
Given a positive test, the chance of having it is 180 out of 260, about 0.69, not 90 percent.
已知阳性时，真正患病的概率是 180/260，约 0.69，而不是 90%。
Check yourself.
自我检测。
40 games at 0.2 expect 8 wins.
40 局、胜率 0.2，预计赢 8 局。
If P of A is 0.35, P of not A is 0.65.
P(A) = 0.35，则 P(非 A) = 0.65。
A coin and a die give 2 times 6, so 12 outcomes.
硬币和骰子共 2 × 6 = 12 种结果。
With replacement, two reds is 3 fifths times 3 fifths: 9 over 25.
放回时两次都红是 3/5 × 3/5 = 9/25。
And given a pupil cycles, the denominator is 18, so swim given cycle is 7 over 18.
已知骑车，分母是 18，所以 P(游泳 | 骑车) = 7/18。