A high school level probability question, a high school math probability problem, please point out

Multiple independent events are probable.

Answer: Each card has a probability of 1 5 on each slot and does not affect each other.

Then the total number of possible situations is 5*5*5*5*5=3125.

One, five are the same.

The suit can only be one color with C5 1 = 5

There is obviously only one way to arrange it: a1 1=1

Result = 5 * 1 3125 = 5 3125 = 1 625 Two and four are the same and one is different.

There will be two kinds of suits: c5 2*c2 1=20

Arrangement = A5 1 = 5

Result = 5 * 20 3125 = 100 3125 = 4 125 3 and 3 are the same The other two are different.

There will be three kinds of suits: C5 3 * C 3 1 = 30 arrangement = A5 1 * A4 1 * A1 1 = 20 = C5 3 * A 2 1 = 20

Results = 600 3125 = 24 125

Fourth, three are the same and the other two are the same.

There will be two kinds of suits: c5 2 *c 2 1=20, and the arrangement will appear a5 2 c2 1==c5 2 = 10, so the result = 200 3125 = 8 125

5. Two are the same and the other three are not the same.

There are four suits: C 5 4 * C 4 1 = 20, and the arrangement will appear A5 5 2 = 60

Results = 1200 3125 = 48 125

Sixth, two pairs of two identical.

There are three suits: c5 3 *c3 1=30

The arrangement will show a5 5 2 2 = 30 and the result = 900 3125 = 36 125

Seven and five are different.

There is only one possibility.

Arrangement: a5 5 = 5 * 4 * 3 * 2 * 1 = 120 result = 120 3125 = 24 625

PS 1: The above equation is calculated step by step to make it easier for LZ to see.

2.One of them doesn't know where you can hi me.

3.The above equation can be seen with all probabilities, and the probability of adding 1-7 equations is 1 to verify each other.

There are only these seven cases.

Each tile has a 1 in 5 chance of being dealt.

One, 5 cards of the same probability 1 5 5 = 1 3125 two, 4 same, 1 different probability 1 5 4 = 1 625 three, 3 same, the other 2 different: 1 5 3 (1-1 5 2) = 24 3125

Fourth, 3 are the same, and the other 2 are the same: 1 5 3 1 5 2 = 1 3125 Five, 2 are the same, and the other 3 are not the same from each other; 1/5^2×(1-1/5^3)=124/3125

Six, two pairs of 2 are the same, but not 4 are the same; It's hard.

Seven, 5 cards are different from each other (1-1 5 5) = 3124 3125

Calculate the possible situations in each ranking first, and then ask for all of them, I won't talk about the specific process because it's too troublesome. In short, this kind of question is calculated like this, first the part and then the whole.

College "Probability Theory and Mathematical Statistics", it seems that there are such questions in high school, but I can't remember clearly.

Upstairs is wrong, and each rank is different.

I:1 625;II: 4 125;Three: 12 125 I tried my best In a few days I will have you another answer Ask for a reward.

After the first 2 steps, the number of white balls in a may be 5, 6, 7. Categorical discussions.

5: The first ball sells silver slag is a white ball, the second ball is a black ball, probability = 6 10 * 5 12 = 1 4. Then the third is the white ball probability = 5 10 * 1 4 = 1 8

7: The first ball is a black ball, the second ball is a white ball, probability = 4 10 * 6 12 = 1 5Then the third is the white ball probability = 7 10 * 1 5 = 7 50

6 pcs.; Probability = 1-1 4-1 5 = 11 20. Then the third is the probability of white ball = 6 10 * 11 20 = 33 100

To sum up, the probability that the 3rd ball is white is 1 8 + 7 50 + 33 100 = 119 200

Do the math yourself! 4 10*6 12*6 10+6 10*7 12*6 10+6 10*5 12*5 10+4 10*6 12*7 10=??

I'll do the math for you.

p(3rd=w) =p(no change)(6 10) +p(swap)(5 bypass 10).

2. First of all, you need to calculate how many coloring schemes there are.

A and D are not adjacent, then they can be of the same color, and they can be discussed as a classification.

1) Same color. A and D have 5 options, C has 4 options, and B and E are different colors from A(D) and C, so they are all 3 options.

Altogether it was 5 4 3 3

2) Different colors.

Think about it for yourself, and the above is the same reason, two plus one, is all the dyeing possibilities.

Then, analyze that A and D are both red.

Then C has 4 options, B and E are not red, and they are not the same color as C, so they are all 3 options.

Therefore, there are 4, 3, 3 possibilities for a and a d to be red.

Then compare before and after, and the answer comes out.

3. At most two of them are red, which may be A, D, B, E

The probability that a and d are red has been calculated, and b and e can also be calculated.

One is red, the same counts, 4 areas.

There is no red, which is equivalent to 4 colors of dyeing, and the results are obtained by continuing to divide A and D into the same color and different colors.

0 red: a, d same color: 4 3 2 2 a, d different color: 4 3 2 1 1 two sums: 72

1 red: a, b, d, e any one is red: 4 3 2 2 c is red: 4 3 3 + 4 3 2 2 The sum is 276