Mathematical Probability Solution, Mathematical Probability Solve Detailed Solution

Empirically, we regard the victory and defeat as a simple comparison of chess strength, without considering the accidental factor, then B has a 0% winning rate, and A wins all.

Mathematically, we regard victory and defeat as a simple probability event, and believe that people have two states, victory meets defeat, defeat meets victory, and the same state meets draw. Then A has a 70%*60%=42% win rate, B has a 40%*30%=12% win rate, and the remaining probability corresponds to a draw, because a draw is possible.

B Khan, this cannot be calculated by simple probability.

Because. A's win rate is 70% because he is better than 70% of people.

B has a 40% chance of winning because he is better than 40% of people.

Then A is very likely to be stronger than B, so B can't keep a 40% win rate in front of A.

In daily life, if two of these people meet, A's win rate is close to 100%.

Quite simply, the probability of A winning is.

A wins and B loses: 42%, B wins and A loses 12%. If a draw is ignored, the probability of A winning is 42% (42%+12%)=

A wins: A wins and B loses.

The probability of a draw is:

The probability of A winning is.

The probability of B winning is.

Do not put back the sampling spike sail = continuous sampling.

1. P=(C42*C42+C43*C41+C44) Two or more red balls in C84: two red and two white, three red and one white brigade belt, and four red.

2. P=(C32*C31+C33) C63 Two or more red balls: two red, one white, three red.

Just enumerate all of them.

Make use of classical generalizations.

1) [40, 45) = people, [45, 50) = people.

Fashionable family) [40, 45) = 150 * 40% = 60 people, [45, 50) = 100 * 30% = 30 people.

2) Stratified sampling: the ratio of fashion families in the two groups is 2:1, so 6 people are selected from [0,45) and 3 people are selected from [45,50), so the distribution is listed as x=0 p=c(3,6) c(3,9)=5 21,x=1,p=c(1,3)c(2,6) c(3,9)=15 28,x=2,p=c(2,3)c(1,6) c(3,9)=3 14,x=3,p=c(3,3) c(3,9)=1 84, So e(x)=0*5 21+1*15 28+2*3 14+3*1 84=1

Another victim of 100 prestigious schools...

Suppose the same dish can only be drawn once. In this case, the sample space is limited, so the answers to each question are different.

2) 40 198 = (Note: two elements are missing from the sample space) 3) 157 198 =793

4）160/200 159/199 158/198 157/197 40/196 = 160p4 * 40/200p5 = .0830

This is not too good.,One because it's tangled and because it's a little difficult.,I don't understand this.。

If it is randomly proportional, it is 1 4 each.

The probability of getting the white ball for the first time is c(3,1) c(5,1)=3 5The probability of getting the white ball for the second time is c(2,1) c(4,1)=2 4=1 2, so the probability of getting the white ball both times is.

The first probability is p=3 5

The second probability is p=2 4

Twice probability p=3 5*2 4=6 20=3 10 (the probability of two independent events is multiplied twice).

Take the ball twice, exactly the probability of getting the white ball.

p=(3/5)*(2/4)=3/10