Khan Public lessons-Probabilistic learning Notes (ii) No sequence independent events, mathematical symbols, Bayes ' s law, unfair probability calculation

independent events with no order

For example 1:flip coins, toss 4 times, ask 2 times for positive probability.

All possible permutations of probability are 2x2x2x2, the number of events that meet the requirements can be listed, but if the number of times to throw more, is impossible, a different way of thinking.

These 2 heads are placed in 4 positions, and there are a number of mid-release methods. Assuming that one proves to be ha, the other is hb,ha can be placed in 4 positions, and when HA is determined, HB can put the remaining 3 positions, so the method 4x3=12. However, in fact, HA and HB are not sequential requirements, that is, HAHBTT and Hbhatt is the same, in the calculation of the total number of permutations are also treated as the same, in addition to the number of repetitions, the total arrangement of HAHB may be 2x1=2. The possible permutations are 12/2=6. The way of thinking is the same as the sequence-dependent event, which is 4 2:

Example 2: Fair coins, Cast 5 times, 3 times positive probability

The general situation is to throw n coins, with k positive probabilities:

some mathematical symbols

The data conforms to the programming symbols of the computer, do not confuse.

For example, in the flip die example, the probability of getting 2 points or 4 points is recorded as: P (2∪4) =p (2) +p (4), which is the symbol of the ampersand, equivalent to "or", and the intersection of the symbol ∩, indicating "and", the Chinese and English in the expression will be different.

For example, Flip 2 dice, add up to 7 points, ask for a 1-point and a 6-point probability, record P (event|conditions) =p (one 1 One 6 | total sum=7), "|" The right side is the specified condition.

Bayes ' s law

example, there are 10 coins,9 are fair coins, one is unfair coin, both sides are positive, we randomly draw a coin, throw 5 times, 5 times are positive, ask what the probability of unfair coin. This example is called: P (Unfair coin | 5 flips all heads), the occurrence of probability under the condition of the record

P (a∩b) =p (a|b) p (b), P (b∩a) =p (B|a) p (a), and P (a∩b) = P (b∩a)

Therefore: P (a|b) p (b) =p (B|a) p (a) –-> this is Bayes ' law (Bayesian rule). This example

Example : There are 5 fair coins,10 unfair coins (head probability 0.8,tail chance 0.2), ask if you pull a coin, throw 6 times, 4 times is head, for this coin is the probability of fair coin.

A:fair Coin, B:4/6heads/flips

P (B|a) =p (6 flips 4 heads| fair coin) = number of combinations x 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

probability calculation of unfair event

The probability of shooting is 80%, the probability of asking 5 to cast 3 is how much.

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

For a 5 cast three, such as BBSSB, the probability is 0.8*0.8*0.2*0.2*0.8=0.83*0.22, for any 5 cast 3 is the same, so there are: P (5 cast 3) =10x0.83*0.22=20.48%

The course behind the probability series is a two-item, Poisson distribution, as already mentioned in the statistical series.

The End.

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