Mathematicians of LRC, help me out. 4 people are racing. They have a 75% chance of success. What are odds that 2, 3 or 4 bomb?

Expert here. For two bombing would be 1/4 × 1/4 which is 1 in 16 chance. Three bombing would be 1/4×1/4×1/4 which is 1 in 64 Chance. All 4 bombing would be 1 in 264 chance.

I may or may not be right. I did have 6 beers today which would justify any incorrect calculation.

ChatGPT is a wordcel that can't do math

Exactly 2 bombing = (1/4)*(1/4)*(3/4)*(3/4)*(4 choose 2) = ~21.1%

Exactly 3 bombing = (1/4)*(1/4)*(1/4)*(3/4)*(4 choose 3) = ~4.7%

4 bombing = (1/4)^4 = ~0.4%

2 or more bombing = 21.1% + 4.7% + 0.4% =~26.2%

Your math is correct for the odds that all four bomb, but I don’t think it is right for the others. I am too stupid to explain my thinking, but, for example, i think it is a 1/16 chance than any two SPECIFIC athletes bomb. But, if the second doesn’t bomb, there is still a1/4 chance that athlete 3 will bomb, and 1/4 for athlete 4. I don’t know the correct answer, but ya

That's it? I guess the day is still young.

First place is out of the equation because they didn't fail. So you're left with 2 out of your three to place since you're racing 4 people. Success is guaranteed for three of them. So your probability is based on a 1/3 chance of success and also a 1/3 chance of failure since only one gets left out, so since 3/4 are in and 1/4 is left out, and 1/3 is left out then it would come to 3/4 *(1/4 +1/3) or 3/4 * 7/12 or 21/48 or 7/16 which is .4375 but you're basing this on 75 percent potential...

So .75 (.4375) equals .3281

Bard seems the most correct

This is a simple binomial calculation:

Image:i.ibb.co/HtxhH6C/image.png

p = probability of success = 0.75
q = probability of failure = 1 - p = 0.25
n = number of trials = 4
x = desired numbers of successes = 1

For the case where exactly one runner succeeds and three fail (P(X = 1)) there's a 4.688% chance.

The number Bard gave you, 5.08% is the combined probability of all failing and three failing, P(X ≤ 1).

Assuming it's independent (one failing doesn't cause others to run slower), you only have 16 different scenarios. Here are all of them. Binomial is correct: 1 4 6 4 1

1. ABCD - all run well

2. aBCD - A runs poorly
3. AbCD
4. ABcD
5. ABCd

6. abCD
7. aBcD
8. aBCd
9 AbCd
10 AbcD
11 ABcd

12-15: like 2-5 but reverse.

16: abcd

So:

2 failing: .75^2 * .25^2 * 6 = 21.09%

3 failling: .75 * .25^3 * 4 = 4.69%

4 failing: .25^4 * 1 = 0.39%

and:

1 failing: 42.18%

0 failing: 31.64%

However, the assumption of independent trials doesn't hold for your experiment.

Also, Bard is complete garbage for math.

I got a 0.73828125 probability that 3 or more run "good" races from this calculator with p=0.75 so there's like a 26.17% chance that 2 or more run a "bad" race.

gigacalculator.com

Binomial Distribution Calculator - Binomial Probability Calculator, Binomial CDF calculations

Georgiev, Georgi Z.

Calculates the probability of an event or a number of events occuring given the probability of an event occuring during a single trial and the number of trials. ➤ Online binomial probability calculator using the Binomial Prob...

Link:www.gigacalculator.com/calculators/bin...

People - like the bots -a re all over the place here. Let's start with a simple question. Is it higher or lower of exactly 25% that 2 or more of them bomb.

I think that's how we should think of it.

assuming you mean each has probability p = .75 of success, ergo 1-p = .25 of failure.

All 4 failing is easiest to calculate, since there's only one way it can happen. .25^4

3 of 4 failing can happen 4c3 = 4!/(3!1!) = 4 ways. Equivalent to how many ways exactly one succeeds. Probably of each such independent result is p(1-p)^3, so probability of any happening is

4(.75)(.25^3)

2 of 4 failing can happen 4c2 = 4!/(2!2!)) = 6 ways. Probability is

6(.25)^2(.75)^2

As these are all independent outcomes, you can add up the results for the total probability.

Might be a typo or error in here somewhere, but it's how to do it. Going backwards (probability of all or 1 succeeding, then subtracting from 1) would be simpler.

Ivy math major, gidbob, and I all posted the same answer

I agree with a previous responder that its a simple binomial distribution.

I'm more concerned that you are saying "please" to a computer.

And it's correct. If rojo knew how to use a binomial calculator he would have it.

Given your sample size, equal .Because three make it. One doesn't.

You have not provided other variables.

Even if 3/4 make it at 75 percent and one bombs, it's still 25 percent because your sample size is 4.

These bots really can't be trusted. I just rephrased it as follows.

4 people are going to the casino. They each have a 25% chance of returning home with zero dollars. What percent of the time will 2 or more of them return home with zero dollars?

Bard said this:
There are several ways to solve this problem. Here's one method:
Calculate the probability of no one returning with zero dollars: This is the opposite of what we're looking for, but it's easier to calculate first. The probability of one person NOT returning with zero dollars is 1 - 0.25 = 0.75. Since the outcomes are independent (their results don't affect each other), the probability of all 4 not returning with zero dollars is 0.75^4 = 0.31640625.
Subtract from 1: Since the two events are complementary (all 4 not losing with 2 or more losing), the probability of 2 or more losing is 1 - 0.31640625 = 0.68359375.
Convert to percentage: Multiply by 100% to express the answer as a percentage: 0.68359375 * 100% ≈ 68.36%.
Therefore, there is approximately a 68.36% chance that 2 or more of the 4 people will return home with zero dollars.

Chatgpt acutally agreed but phrased it differently.

To solve this problem, we can use the complement rule. The probability of at least 2 people returning home with zero dollars is the complement of the probability that none or only one person returns home with zero dollars.
Let's calculate the probability that none or only one person returns home with zero dollars:
Probability that all four return home with money:
�(All return with money)=(0.75)4P(All return with money)=(0.75)4
Probability that exactly one returns home with zero dollars (and the other three return with money):
�(One returns with zero dollars)=(41)×(0.25)1×(0.75)3P(One returns with zero dollars)=(14)×(0.25)1×(0.75)3
Now, let's find the combined probability of these two events:
�(None or only one returns with zero dollars)=�(All return with money)+�(One returns with zero dollars)P(None or only one returns with zero dollars)=P(All return with money)+P(One returns with zero dollars)
�(None or only one returns with zero dollars)=(0.75)4+4×0.25×(0.75)3P(None or only one returns with zero dollars)=(0.75)4+4×0.25×(0.75)3
Now, the probability that 2 or more return home with zero dollars is the complement:
�(2 or more return with zero dollars)=1−�(None or only one returns with zero dollars)P(2 or more return with zero dollars)=1−P(None or only one returns with zero dollars)
Let's calculate this:
�(2 or more return with zero dollars)=1−[(0.75)4+4×0.25×(0.75)3]P(2 or more return with zero dollars)=1−[(0.75)4+4×0.25×(0.75)3]
�(2 or more return with zero dollars)≈1−0.68359375P(2 or more return with zero dollars)≈1−0.68359375
�(2 or more return with zero dollars)≈0.31640625P(2 or more return with zero dollars)≈0.31640625
So, approximately 31.64% of the time, 2 or more out of the 4 people will return home with zero dollars.