Ask a question about calculating probability, thanks

Answer: This is the calculation method of c subscript 16 superscript 3 in the combination problem.

c Subscript 16 Superscript = 16!/3!*(16-3)!

If you choose 16 people, if each person has a number (there is an order), it is a matter of arrangement, which is equal to 16! But now there is no order, it is a combination problem, and the order must be removed.

After selecting 3 people, how many of them are likely to be in order? The answer is: 3!*(16-3)!

Why? Because the 3 people selected are: 3! There are 13 of the remaining 13 people!

multiply the two, i.e.: 3! *（16-3）！

This is a basic question, but it is a bit difficult to understand when you are a beginner in permutations.

I was confused back then, too.

I hope mine can help you.

I don't understand your calculations, but I'll make an analogy from another side and you'll understand.

The probability calculation is the same as for the Double Color Ball lottery.

There are 16 probabilities for choosing the first person, and there are only 15 probabilities for choosing the second person, and the third one is 14 probabilities. So the calculation is 16*15*14=3360 combinations.

The answer is that the numerator and denominator are reversed, and this is C-16-3, so to speak, 16 people randomly arranged in a team (16!).Take the first three people (3!Because it's a random line, the way to take these three people is to take three out of 16 people, but after taking three, the next 13 people have (13!).

The situation is double-counted (that is, the first three people take the last 13 and there are so many arrangements) and finally 16*15*14 3*2*1

It's a combination of problems, c16 (3) is to randomly select 3 people from 16 people.

Summary. Please describe your problem more specifically, you can use ** and other forms, and talk to the teacher in detail, so that the teacher can better help you.

Please describe your problem more specifically, you can use ** and other forms, and talk to the teacher in detail, so that the teacher can better help you.

If the probability of winning a thing is 10%, what is the probability of winning it ten times in a row, and what is the specific formula?

What if it's 30% five times in a row?

The same thing is true.

It's not the same if you don't win the lottery in a row.

Let these 5 barrels be 1, 2, 3, 4, 5 respectively, if Na Xun throws 5 balls in a row and hits the No. 1 bucket, then the general Chang ride rate is (1 5) * (1 5) * (1 5) * (1 5) * (1 5) = 1 3125, and five consecutive hits to the same barrel have five barrels of five situations, so the probability of throwing 5 balls in a row to hit the same barrel is (1 3125) * 5 = 1 625

It's a step-by-step problem, so it should be multiplication.

The probability of each ball thrown into the first bucket is 1 5, so the probability of throwing the first bucket together should be 1 5 to the fifth power.

This is a classic classical probability problem. If you must enter the bucket every time to search for the first orange, then one ball per group is equivalent to a one in five probability of entering a certain bucket. And each ball into a barrel is a separate event, so the probability of five consecutive balls into a barrel is.

The probability of hitting a bucket in a row is a sharp one.

1 5 1 5 1 5 1 5 1 5 1 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 1 5 1 5 1 5 1 5 1 5 1 5 1

The total number of cakes is 6 50 + 21 = 321, the probability of a single piece of paragraph 1 to paragraph 6 is 50/321, and the probability of three times is 150/321; The probability of a single insertion of paragraph 7 is 7 per 107, and the probability of three insertions is 21 per 107;

Changed to a percentage, paragraph 1 to paragraph 6, the probability of putting each bag is about; The probability of putting each bag of paragraph 7 is about.

1. The probability that the evaluation is good = 60% * 70% + 40% * 20% = 50% 2, if the evaluation of a commercial building is good, let it be successful The probability is: 50% * x (good evaluation and success) = 60% * 70% (evaluation is good and successful) x = 84%.

for (1 2) n

aa is (1-(1 2) n) 2

Probability (1-(1 2) n) (1+(1 2) n)2Start: aa=1 aa=0 aa=0

> the first generation: aa=1 4 aa=1 2 aa=1 4, then this generation produces a as (1 4 + 1 2 2) = 1 2 produces a as (1 4 + 1 2 2) = 1 2

The next generation of AA has (1 2) 2=1 4 AA has 2*(1 2)*(1 2)=1 2

aa has (1 2) 2 = 1 4

So it's still 1 4:1 2:1 4

3.The same (because no matter how many inbred deaths, the total number of dominant homozygosity remains the same) 4Start the first generation as above.

For aa: 1 4, aa: 1 2 aa: 1 4aa, all dead: aa: 1 3 aa: 2 3

The probability of producing a is p1 and the probability of a is q1

p1=(aa+aa/2)=1/3+1/3=2/3q1=(aa/2)=1/3

So aa:p0 2=4 9 aa:2p0q0=4 9 aa:q0 2=1 9

aa is all dead aa:aa=1:1

If you continue to do so, it is equivalent to not looking at AA, only AA and AA are in P2=(P1 2+P1Q1 2) Q2=(P1Q1 2)AA2 The second is P2 2 AA is 2P2Q2 Probability AA:P2 2 (P2 2+2P2Q2).

aa:(2p2q2) (p2 2+2p2q2) I don't know how to count this sequence.

The probability of a bulb with a life of more than 1,000 hours should be calculated, if there are 1,000 bulbs, then there are 600 bulbs with a life of more than 1,000 hours, and 400 with a life of more than 1,500 hours, so the probability should be.

The effect of drawing 3 cards sequentially and 3 cards at once is equivalent, so this is a hypergeometric distribution. Then the probability of drawing the correct hole card is c(2,2)*c(8,1) c(8,3)=(1*8) 56=1 7