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Probability problem of coin toss
Your algorithm is obviously wrong.

Your 1/32 should represent the probability of consecutive heads when you flip a coin five times. However, "toss a coin 10 times, of which at least 5 times are one-on-one hit" does not require the first 5 consecutive one-on-one hits.

1, an event with equal probability is an event with equal opportunity. For example, if you toss a coin at random, the chances of heads or tails are 50/50.

2. Events with unequal probabilities have unequal probabilities. For example, when a coin is tossed ten times, there are 1 1 heads: 0 heads 10 tails, 1 heads with 9 tails and 2 heads with 8 tails. . . . . 10 is positive and 0 is negative. However, these combinations are not of equal probability, so it can't be said that there are 6 kinds with more than 5 times, with a total of 1 1, and the probability is 6/1. This is not right.

3. When calculating the probability, we should use "equal probability events" for comparison. "Unequal probability events" are more complicated.

The above is the explanation, and the following is the official start. Because the target is junior high school students, I will talk about it in detail (I happen to be a high school teacher).

1. Suppose we throw 10 coins in a row. How many arrangements are there?

The power of 2 is 10. This is all possible arrangements.

2. Because every time a coin is tossed, both sides are equally probable, so the arrangement of "10 to the second power" is equally probable.

This is why we should use permutation, not combination. The combination is not equal probability. (As mentioned above)

3. In all these arrangements.

There are 65,438+00 possible arrangements for face up: 65,438+0 (mathematical expression: C 10 (10)).

There are nine possible permutations of the head: 10 (mathematical expression: C 10(9)).

Explain here. Suppose there are ten empty seats in the row in front of you. You have to put nine heads and the rest with tails. How many choices do you have? 10. It is equivalent to choosing 9 out of 65,438+00 spaces to put coins on the front. Because in mathematics, this behavior of "10 choosing 9" has 10 possibilities, which is the meaning represented by C 10(9). (This is a formula that cannot be entered in Baidu. Correct is that 10 to the right of C is at the bottom and 9 is at the top. )

There are 8 possible arrangements in head-up: 45 (mathematical expression: C 10(8)).

In the same principle as above.

There are seven possible arrangements for face up: 120 (mathematical expression: C 10(7)).

There are six possible arrangements for head-up: 2 10 (mathematical expression: C 10(6)).

There are five possible arrangements for looking up: 252 (mathematical expression: C 10(5)).

Therefore, there are at least five possible head-up arrangements:

c 10( 10)+c 10(9)+c 10(8)+c 10(7)+c 10(6)+c 10(5)

And all permutation numbers are 2 10 power = 1024.

Therefore, the probability of five heads-up shows the proportion of "five possible heads-up arrangements" in "all arrangements".

The probability of five heads-up appearances =638/ 1024=63.2% (consistent with the answers of the above two guys).

Generally speaking, the probability is over 60%.

You can verify that throwing 10 coins at random counts as a group. Do more groups.

At least five heads must be the majority. Not the small probability of 1/32 you mentioned first.

By the time I answered, the above two guys had got it right. Although you look a little different from my algorithm, it's actually the same thing. I just said it in detail.

Ps: Probability theory is a very interesting thing. Unlike other branches of mathematics, it is so easy to solve by calculus and drawing. Most of the time, I think in my head. I think it's very simple, but I don't understand it. I don't know what's going on if I give you the answer.

I hope I can think more and have my own experience.