Probability of getting 3 out of 4 numbers correct

• Phys12
In summary, the probability of winning the small prize in a lottery by matching three digits correctly is 36/10000, or 0.0036. This can also be calculated by considering the ways of getting three digits correct, which is equivalent to choosing 3 out of 4 digits correctly. This can be represented as 4 pick 3, or 4 choose 3, which is equal to 4. Thus, the probability is also 9/10000 multiplied by 4, which gives us the same result of 36/10000 or 0.0036.
Phys12
Problem statement: In a lottery, players win a large prize when they pick four digits that match, in the correct order,four digits selected by a random mechanical process. A smaller prize is won if only three digits are matched. What is the probability that a player wins the small prize?

Solution: The total number of choosing 4 digits are: 10 * 10 * 10 * 10 = 10,000. Ways of choosing three digits correctly out of 4 is the same choosing one incorrectly. Ways of choosing the first digit incorrectly are 9, same for second, third and fourth. You add the total number of ways to get 9 + 9 + 9 + 9 = 36 and then divide by the total number of ways of choosing 4 digits to get the probability: 36/10000

How would you solve this problem by not considering the different ways of getting the digit incorrect, but by considering the ways of getting three digits correct?

My attempt: If I think about getting the digits correct, then for getting the first three digits correctly, I have 9 ways of picking it since the 4th digit can be any digit except for the correct one, similarly for the other three digits and another and then one more time. Giving us again 9 + 9 + 9 + 9 = 36 ways. Is that a/the correct way of thinking about it?

Think of this as four different trials. The odds of being successful in each trial are 0.1 or 10%. Now you want the probability of being successful in 3 but unsuccessful in one. So for example if you are successful in the first three trials but then fail in the last you have such a condition, and the odds of this happening are ##0.1\cdot 0.1 \cdot 0.1 \cdot 0.9 = \frac{9}{10000}##. But you could also avail in the first two, then fail the 3rd test and then again be successful in the last, and that would also meet your conditions, the odds of it happening would again be ##0.1\cdot 0.1 \cdot 0.9 \cdot 0.1 = \frac{9}{10000}##. So how many ways are there of this happening? Well you have four trials and you want to pick 3 to succeed in (or equivalently, 1 to fail in) so that's 4 pick 3, which is the same as 4 pick 1, i.e. we want ##\binom{4}{3} = \binom{4}{1}=4##.

So the total chances of the event that you are successful in 3 trials and unsuccessful in the other are ##\frac{9}{10000} \binom{4}{3} = \frac{36}{10000}## as you calculated. I don't know if this was the thought process you were describing but I believe it's a pretty general way to think about it.

edit: If you have not seen this "4 pick 1" function before but are interested, I'm more than happy to discuss it.

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PeroK and Phys12
GwtBc said:
edit: If you have not seen this "4 pick 1" function before but are interested, I'm more than happy to discuss it.
I know what it is, I don't understand its application in this question. The "4 pick 1" function, as I understand it, is equivalent to saying that there are 4 choices, in how many ways can I pick 1 of them, if I do not care about the order. In this question, out of the 4 digits, we pick three of them, and that's (4 3) and the odds of being successful in each pick is 9/10^4, so we multiply the two. Is that correct?

GwtBc
Phys12 said:
I know what it is, I don't understand its application in this question. The "4 pick 1" function, as I understand it, is equivalent to saying that there are 4 choices, in how many ways can I pick 1 of them, if I do not care about the order. In this question, out of the 4 digits, we pick three of them, and that's (4 3) and the odds of being successful in each pick is 9/10^4, so we multiply the two. Is that correct?
Another description of ##\binom{n}{k}## is:

If you have ##n## things and ##k## have a certain property, then that's the number of ways you can do it.

In this case, you have 4 things and you want 1 to be wrong and 3 to be right, then that is ##\binom41##, which is equal to ##\binom43##.

GwtBc
Phys12 said:
I know what it is, I don't understand its application in this question. The "4 pick 1" function, as I understand it, is equivalent to saying that there are 4 choices, in how many ways can I pick 1 of them, if I do not care about the order. In this question, out of the 4 digits, we pick three of them, and that's (4 3) and the odds of being successful in each pick is 9/10^4, so we multiply the two. Is that correct?
Correct!

Phys12 said:
Problem statement: In a lottery, players win a large prize when they pick four digits that match, in the correct order,four digits selected by a random mechanical process. A smaller prize is won if only three digits are matched. What is the probability that a player wins the small prize?

Solution: The total number of choosing 4 digits are: 10 * 10 * 10 * 10 = 10,000. Ways of choosing three digits correctly out of 4 is the same choosing one incorrectly. Ways of choosing the first digit incorrectly are 9, same for second, third and fourth. You add the total number of ways to get 9 + 9 + 9 + 9 = 36 and then divide by the total number of ways of choosing 4 digits to get the probability: 36/10000

How would you solve this problem by not considering the different ways of getting the digit incorrect, but by considering the ways of getting three digits correct?

My attempt: If I think about getting the digits correct, then for getting the first three digits correctly, I have 9 ways of picking it since the 4th digit can be any digit except for the correct one, similarly for the other three digits and another and then one more time. Giving us again 9 + 9 + 9 + 9 = 36 ways. Is that a/the correct way of thinking about it?

I think your solution ignores the problem stipulation "...digits that match, in the correct order, ..." The problem seems to be saying that order is important, so we need to be looking at permutations, not combinations. Assuming that the ten digits 0,1,...,9 can each be chosen only once, the total number of possibilities is ##N = 10 \times 9 \times 8 \times 7 = 5040.##

Presumably, if the lottery selects ##abcd## you will a smaller prize if you pick either ##abcX## with ##X \neq d## or ##Xbcd## with ##X \neq a##. What is not clear (and may vary according to Lottery rules) is whether ##aXcd## with (##X \neq b##) or ##abXd## (with ##X \neq c##) could also win a small prize; they have three of the correct letters with ##a## before ##b## before ##c## before ##d,## but they do not occur in uninterrupted strings. Maybe that is not good enough---or maybe it is. If unsure, why not solve the problem both ways and explain the assumptions underlying each way?

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What is the probability of getting exactly 3 out of 4 numbers correct?

The probability of getting exactly 3 out of 4 numbers correct depends on the context and the method of selecting the numbers. In general, if you are randomly selecting 4 numbers from a set of n numbers, the probability would be 3/n. However, if the numbers are being selected without replacement, the probability would be slightly different. It is always important to specify the context in order to accurately calculate the probability.

How is the probability of getting 3 out of 4 numbers correct calculated?

The probability of getting 3 out of 4 numbers correct can be calculated using the formula for combinations. This formula takes into account the number of ways to choose 3 numbers out of 4 and the total number of possible outcomes. The formula is: P(3 out of 4 correct) = (number of ways to choose 3 numbers out of 4) / (total number of possible outcomes).

Is the probability of getting 3 out of 4 numbers correct the same as getting 3 specific numbers correct?

No, the probability of getting 3 out of 4 numbers correct is not the same as getting 3 specific numbers correct. In the first case, you are selecting any 3 numbers out of 4, while in the second case, you are selecting 3 specific numbers out of 4. The probability will be different in these two scenarios as the number of possible outcomes will vary.

Can the probability of getting 3 out of 4 numbers correct be greater than 1?

No, the probability of getting 3 out of 4 numbers correct cannot be greater than 1. In probability theory, the maximum probability that can be assigned to an event is 1, which represents a 100% chance of the event occurring. Any probability greater than 1 is not possible and indicates an error in calculation.

How can the probability of getting 3 out of 4 numbers correct be used in real-world applications?

The probability of getting 3 out of 4 numbers correct can be used in various real-world applications such as in gambling, lottery, and statistical analysis. It can also be used in predicting the likelihood of success in certain scenarios where the outcome depends on choosing a specific number or combination of numbers. However, it is important to note that the probability is not a guarantee of the outcome and should be used as a tool for making informed decisions.

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