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).
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..?
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?
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.
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 ?
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?
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
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! :)
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)
All over the internet and you are they only one who showed how to do this correctly
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).
This is awesome! so easy to understand the way you explained it. Thanks!
I can't stop seeing a face on the right of the board
astonishing content Center of Math. I crushed that thumbs up on your video. Always keep up the solid work.
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..?
This video just saved my life
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?
can you rewrite the last formula please (at least x ) one
THANK YOU! currently studying for my statistics final ahahaha
Why did u multiply with 3/6 and it was probability of all men ? I didn't get that point..
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.
thanks bro
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?
thank you so much.
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 ?
Yes
Thank u
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?
That is correct!
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.
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
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! :)
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)
Great...u sound like Robert Downey Jr
Concentrate on studies bro
matay di maklaro ang sa atleast 2
God bless lmao
u look like zukerberg