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Tossing coins - calculating probabilities

As a preface. Everyone knows that a coin has two sides - heads and tails. Numismatists believe that the coin has three sides - obverse, reverse and edge. And among those, and among others, few people know what a symmetrical or mathematical coin is. But they know about it (well, or should know :), those who are preparing to take the exam. In general, this article will focus on an unusual coin, which has nothing to do with numismatics, but, at the same time, is the most popular coin among schoolchildren. Tossing coinSo. A symmetrical coin is an imaginary mathematically perfect coin with no size, weight, or diameter. As a result, such a coin also does not have a edge, that is, it really has only two sides. The main property of a symmetrical coin is that under such conditions the probability of falling heads or tails is exactly the same. And they came up with a symmetrical mathematical coin for conducting thought experiments. The most popular problem with a mathematical coin sounds like this - "In a random experiment, a symmetrical coin is tossed twice (three times, four times, etc.). Find the probability that one of the sides will fall out a certain number of times."

Solving the problem with a Tossing Coin

It is clear that as a result of tossing the coin will fall either heads or tails. How many times - depends on how many throws to make. The probability of getting heads or tails is calculated by dividing the number of outcomes that satisfy the condition by the total number of possible outcomes. Consider the solution of this problem on specific examples.

In a random experiment, a coin is tossed once

Everything is simple here. Either heads or tails will come up. That is, we have two possible outcomes, one of which satisfies us.

 - 1/2=50%

 

In a random experiment, a coin is tossed twice

For two throws can fall:

So there are only four possibilities. Problems with more than one throw are easiest to solve by making a table of possible options. For simplicity, let's denote heads as "0"and tails as "1". Then the table of possible outcomes will look like this:

00
01
10
11

If, for example, you need to find the probability that heads will fall once, you just need to count the number of suitable options in the table - that is, those rows where heads occur once. There are two such lines (second and third). So the probability of getting one heads in two tosses of a symmetrical coin is 2/4=50%
The probability that the heads in two throws will fall out twice is 1/4=25%, since two eagles appear in the table once (first line).

 

In a random experiment, a coin is tossed three times

Compiling a table of options:
000
001
010
011
100
101
110
111

Those who are familiar with binary calculus understand what we have come to. :) Yes, they are binary numbers from "0"to "7". This makes it easier not to get confused with the options, since the rows of the table of options are a logical sequence. Let's solve the problem from the previous paragraph - we calculate the probability that the eagle will fall out once. There are three lines where "0"occurs once. This means that the probability of getting one eagle in three tosses of a symmetrical coin is equal to three out of eight - 3/8=37.5%

The probability that heads in three throws will fall out twice is also equal to 3/8=37.5%, that is, absolutely the same.

The probability that the head in three throws will fall out three times is 1/8 = 12.5%.

 

In a random experiment, a coin is tossed four times

Compiling a table of options:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

The probability that heads comes up once. There are only three rows where "0"occurs once, just as in the case of three throws. But, there are already sixteen options. This means that the probability of getting one eagle in four tosses of a symmetrical coin is three out of sixteen - 3/16=18.75%

The probability of getting heads twice in three tosses is 6/8=75%.

The probability that heads will come up three times in three tosses is 4/8=50%.

 

A coin is tossed more than four times

With an increase in the number of throws, the principle of solving the problem does not change at all - only, in an appropriate progression, the number of options increases. The principle is the same - we compile a table of options and count the number of required results. By dividing the number of results that satisfy us by the total number of attempts, we get the probability of getting the desired result.  

Euro coinage