Khan Open Course-probabilistic learning notes (2) no sequential Independent Events, mathematical symbols, Bayes's Law, unfair Probability Calculation

Unordered Independent Events

Example 1:Flip coins, throwing four times, returns the probability that two times are positive.

The probability of all possible arrangements is 2 × 2 × 2 × 2. The number of events that meet the requirements can be listed. However, if the number of throwing events is large, it is impossible.

The two sides are placed in four positions. Assume that one is Ha, the other is Hb, and HA can be placed in four positions. When ha is determined, Hb can be placed in the remaining three positions, so the placement method is 4 × 3 = 12. However, in fact, HA and HB do not need to be ordered, that is, hahbtt and hbhatt are the same, and they are also considered the same when calculating the total number of permutation. In addition to the number of repeated computations, the total hahb sorting may be 2 × 1 = 2. The possible order is 12/2 = 6. The way of thinking is the same as that of a non-sequential event, that is, select 4 or 2:

Example 2: Fair coins. If it is left five times, the probability that three times are positive is returned.

In general, throwing n coins has k Positive probabilities:

Some mathematical symbols

Do not confuse data with computer programming symbols.

For example, in the flip die example, the probability of 2 or 4 points is calculated as: P (2 then 4) = P (2) + P (4 ), this is the Union symbol, which is equivalent to "or", and the intersection symbol is "and", which is different in Chinese and English.

Another example is flip 2 dice, which adds up to seven points. The probability of one point and one six points is obtained, and P (event | conditions) is recorded) = P (one 1 one 6 | total sum = 7), "|" is the specified condition on the right.

Bayes's Law

In this example, there are 10 coins, 9 are fair coins, and 1 is unfair coin, both sides are positive. We randomly select a coin from it, five times, and five times are positive, q: What is the probability of unfair coin? In this example, P (unfair coin | 5 flips all heads) is used to calculate the probability of occurrence under conditions.

P (A ∩ B) = P (A | B) P (B), P (B ∩ A) = P (B | A) P (), P (A then B) = P (B then)

Therefore, P (A | B) P (B) = P (B | A) P (a) -->, Which is Bayes's law (Bayesian Law ). This example

Example: Five fair coins and ten unfair coins (Head rate: 0.8; Tail rate: 0.2). If you pull a coin, you can drop it for 6 times, and four times you get Head, what is the probability that the coin is fair coin?

A: fair coin, B: 4/6 Heads/flips

P (B | a) = P (6 flips 4 heads | fair coin) = number of combinations × probability of each combination = 6C4 * (0.5) 6 = 0.234375

P (a) = P (fair coin) = 5/15 = 1/3

P (B) = P (6 flips 4 heads) = P (6 flips 4 heads | fair coin) P (fair coin) + P (6 flips 4 heads | unfaircoin) P (unfair coin) = 0.234375*1/3 + 6C4 * (0.8) 4 (0.2) 2*2/3 = 0.241965

Unfair event Probability Calculation

The shooting probability is 80%. Ask the number of the three shots in five.

The first among the third, and the third among the first, there is no difference, so the combination problem. Among all the permutation and combination, the number of combinations in the 5 cast and 3 is:

For a certain five-vote medium, such as BBSSB, the probability is 0.8*0.8*0.2*0.2*0.8 = 0.83*0.22. It is the same for any five-vote 3, so there are: P (5 in 3) = 10 × 0.83*0.22 = 20.48%

The next course of the probability series is the binary and Poisson distribution, which has been discussed in the statistical series.

The End!

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