A high score is used to solve a probability problem, and a high score is used to solve a probability

These are independent events and have no effect on each other, no matter how many times they occur.

Because the first generation of a is known, it is an event that has already occurred, and subsequent events that have not occurred have nothing to do with it. So no matter how many times you spawn a, the probability of producing a is 1 12, and the probability of not producing a is 11 12.

To put it simply, "The first time a is generated, ask what is the probability of a second generation of a" The answer is 1 12

If the question is "what is the probability that the first time a will be produced and the second time a will also be produced", the answer is 1 144

Because each time is random, the first generation of a has no effect on the second and third times, so the probability of the second generation of a is 1 12, and the probability of the third generation of a and not the production of a is 1 12, 11 12

Since it's random.,It's put back after each draw.,So,The probability of the first and second draws is the same 1 12.,If you draw a twice in a row.,It's the product.。。。 If you don't draw it, it's 11 12.。。。 It's been a long time since I've done something probability.,Forget about it.,If it's wrong, I hope you don't laugh at me.。。。

lz I tried my best, I can't recall, it's been a long time since I've covered this knowledge...

This is similar to the probability of a coin toss, and the probability of each time has nothing to do with other cases.

So the probability of producing a for both the second and third times is 1 12

The probability of not producing an a is 11 12 for both of them

The probability of producing an a is 1 12, regardless of the number of times

The probability of not producing an A is 11 12

The probability of producing a for both the second and third times is 1 12

The probability of producing a for both the second and third times is 11 12

It's simple, don't overcomplicate it.

Approximately the same as the above is the calculation process.

Because there is one of the few opposing events that is buried by defective products: there is no defective product.

To be continued. You're thinking correctly, but you're miscalculated.

Calculated to be consistent with the above answers.

For reference, please smile.

For a typical hypergeometric distribution, you can think of 3 of them as **, and then subtract this probability from 1 to get the result.

Your cousin is a little unclear, 1) the probability of winning the lottery three times in a row:

2) Three hidden several times can win the lottery (excluding three times do not win): =

Teacher's solution: A a c B b c a

So B has a 3 in 1 chance of getting b, which is not true. This is equivalent to buying a lottery ticket, and the result can only be two scenarios: winning or not winning.

Is the probability of one in two? The key is not only to look at the result, but also to see what the probability of producing this result is! The probability of B taking (b c) is 1 2, and the probability of taking b in (b c) is also 1 2, so the probability of taking c in the end is 1 2 1 2 = 1 4.

Instead of B, you can only take b c a (three possibilities in total), so the odds of getting b are 1 3. Because the probability of taking b or c is 1 2, the probability of taking a is 1 2, and then in b c with a probability of 1 2, the probability of taking c is 1 2 and the probability of taking c is 1 2 1 2 is 1 2 1 2 .

My classmates did the right thing!

The probability of B taking B is 1 4, because A can choose from A and C when taking the first gift, and the probability of choosing A and C is 1 2, the second step, if A chooses C, then B is A, C is B, if A chooses A, then B can choose B, and the probability of C is 1 2 each, in the previous case B does not get B, do not calculate the probability, then B gets the probability of 1 2 * 1 2 = 1 4.

The probability that A gets c, B gets a, C gets b is 1 2, A gets a, B gets b, C gets C has a probability of 1 4, A gets a, B gets c, and C gets B The probability is 1 4,

Take 6 pairs of 1 color and 12 of them, and then take 1 of the other 2 colors, and finally take 1 of them

So it's 12+2+1=15

Hello! First of all, it is necessary to assume that the occurrence probability of dustpans and snails is the same, which is 1 2

The probability of encountering 1 person of the opposite sex and 20 dustpans is (1 2) 20, and the probability of not meeting Li Zhaoshi is 1-(1 2) 20

Among the n opposite sexes, the probability of not meeting each other in a row is (1-(1 which quarrel 2) 20) n, and the probability of encountering can be 1-(1-(1 2) 20) n

So the probability is related to the total number of people of the opposite sex he meets in his life.

If it helps you, I hope you guess.

The numerator means to take any one of the 6 correct numbers.