Probability: "At Least" Probabilities

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Күн бұрын

This is the fourth video of a series from the Worldwide Center of Mathematics explaining the basics of probability. This video deals with calculating probabilities of "at least 1" or other "at least" probability calculations. For more math videos, visit our channel or go to www.centerofmath.org.

Пікірлер: 29

@kennethrosario3675 8 жыл бұрын

All over the internet and you are they only one who showed how to do this correctly

@zanido9073 2 жыл бұрын

This would've been a lot more useful if you gave an example of at least 2 or more. Finding p(0) is easy, so finding p(at least 1) is easy. But I don't know how to find p(1), which is required to find p(at least 2).

@KeyserTheRedBeard 2 жыл бұрын

astonishing content Center of Math. I crushed that thumbs up on your video. Always keep up the solid work.

@pahulchugh1069 3 жыл бұрын

This is awesome! so easy to understand the way you explained it. Thanks!

@yankhocolleta 6 жыл бұрын

thank you so much.

@godfathergamingyt6466 6 жыл бұрын

can you rewrite the last formula please (at least x ) one

@SolntsaSvet 3 жыл бұрын

I'm having a headache with probabilities here.. I know this video is old and I doubt anyone will check this comment out any time soon (ever?), but I will leave it here, in the hope that someone reads it and might possibly help me out. The issue is that I am getting two different results while trying to solve a problem of "at least" probability using 2 different formulas. I want to calculate the probability P(x) that at least 1 success (x) of an independent event with a probability (p) of 50% occurs in 2 (n) events/attempts. So I calculate it as 1-(1-p)squared, which gives me a P(x)= 1-(0,5x0,5)= 0,75 (75%). On the other hand, when using a binomial distribution formula applied to this example, P(x)= nCx * P to the x * Q to the (n-x), where nCx = n!/(n-x)!*x! I get a completely different result of 0,5 (50%). Something is clearly wrong. I've checked it several times to see if I missed something, but I still get the same results with the binomial distribution formula. So my question is: does the binomial only work/serve, when your x is greater than 1, did i make some stupid mistake in my calculations or there is something essential/fundamental I'm simply failing to grasp here..?

@edgarfranco2449 4 жыл бұрын

This video just saved my life

@94D33M 8 жыл бұрын

thanks bro

@garimachaudhary156 3 жыл бұрын

How do I calculate this; if 82% of students chose maths what is the probability that atleast 8 students chose maths out of 11 randomly selected students?

@isardadi5609 4 жыл бұрын

So if I had to calculate the probability the probability of at least 2 girls out of 6 children (if the probability for boy and girl was the same), my calculations would be 1-((1/64)+(6/64)) = .890625 = 89%, would this be correct ?

@icewater2762 11 ай бұрын

Yes

@abc01234100 2 жыл бұрын

I need help 6% chance, with total 28 times What's the probability of the event happen at least 11 times in all 28 times?

@katewhitmore9755 Жыл бұрын

THANK YOU! currently studying for my statistics final ahahaha

@huskaomar508 6 жыл бұрын

Why did u multiply with 3/6 and it was probability of all men ? I didn't get that point..

@centerofmath 6 жыл бұрын

Hi Huska, To get the probability that from a group of 5 men and 3 women we pick a group of 3 men, we multiply the probabilities that in our first choice of a group member we picked a man, that in our second choice we picked a man, and in our third choice we also picked a man. In our first choice we have 5 men in a group of 8 people, so the probability of picking a man is 5/8. In our second choice we now have 4 men in the group of 7 people (since one man was removed due to our previous choice), so now the probability of picking a man is 4/7. In our last choice we now only have 3 men in a group of 6 people, and so in our last choice the probability of picking a man is 3/6. Multiplying these all together, we get probability of all men = (5/8)*(4/7)*(3/6) = 5/28.

@pmat2757 3 жыл бұрын

Thank u

@PotatoChips-jy9pk 9 ай бұрын

I can't stop seeing a face on the right of the board

@alexismandelias 6 жыл бұрын

So let's say I want to find the probability of the event that: Out of 4 tosses of a fair coin _at least 2_ are tails. I thought that P(at least 2 tails)=1- [P(0 tails) + P(1 tail)] = 1- [1/16 + 1/4] = 1 - 5/16 = 11/16 Is that correct?

@centerofmath 6 жыл бұрын

That is correct!

@alexismandelias 6 жыл бұрын

Great! But how can you compute the P(1 tail)? I did it the long way with a tree-diagram, then counted all possible outcomes.

@centerofmath 6 жыл бұрын

Since each coin flip has 2 possible outcomes, flipping 4 coins has 2x2x2x2 = 16 possible outcomes. To get one tail there are 4 possible outcomes: THHH, HTHH, HHTH, HHHT. So the P( 1 Tail) = 4/16 = 1/4

@alexismandelias 6 жыл бұрын

ooh yes, of course, of course. So basically you have to sort of write down all possible combinations containing one T. I was wondering if there was a more rigorous way to compute this. For example, if you want two tails there is: TTHH, THTH, THHT, HTTH, HTHT, HHTT. so 6/16 = 3/8. Yeah that works for me. Thanks a ton! :)

@centerofmath 6 жыл бұрын

Instead of writing out all possible outcomes, you can also use binomial distribution: www.onlinemathlearning.com/binomial-distribution.html This formula will be easier to use when you have a large number of trials (ex: it would take too long to write out all possible outcomes if you flipped the coin 15 times)