probability of a number occurring in a combination

[HT1992]

i came upon this problem in my work i am not a mathematician i want to know if i choose 5 numbers from 1 to 50 what is the probability of at least one of the being between 41 and 50 (or any 10 numbers in the group)
thank you in advance and sorry if its very simple.

 

[Denis]

...at least one of the being between 41 and 50

Do you mean 42,43,44,45,46,47,48,49?

 

[Romsek]

i came upon this problem in my work i am not a mathematician i want to know if i choose 5 numbers from 1 to 50 what is the probability of at least one of the being between 41 and 50 (or any 10 numbers in the group)
thank you in advance and sorry if its very simple.

The number of of numbers you choose, \(\displaystyle n\), out of 50 being between 41 and 50 is a random variable with a binomial distribution

The probability of choosing a single number from 50 that is between 41 and 50 is simply

\(\displaystyle \dfrac{\text{count of numbers 41-50}}{50} = \dfrac{10}{50} = \dfrac 1 5\)

Thus our binomial distribution has parameters \(\displaystyle n=50,~p=\dfrac 1 5\)

You want to know \(\displaystyle P[n \geq 1] = 1 - P[n=0]\)

The binomial distribution is

\(\displaystyle P[k] = \dbinom{n}{k}p^k (1-p)^{n-k},~k \in 0,1,\dots, n\)

I leave it to you to plug the numbers in and finish

You may be unfamiliar with \(\displaystyle \dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}\)

I suppose you may also be unfamiliar with \(\displaystyle n! = n \cdot (n-1) \cdot (n-2) \cdot \dots 3 \cdot 2\)

 

[Romsek]

Do you mean 42,43,44,45,46,47,48,49?

he says "or any 10 numbers in the group" so I'm going with 41,42,43, ... 49, 50

 

[Jomo]

Do you mean 42,43,44,45,46,47,48,49?

My poor Denis, I worked with you for hours on how to count. Did you learn nothing? The list of number you wrote only contains 8 numbers while the OP wants 10 numbers.

 

[pka]

i came upon this problem in my work i am not a mathematician i want to know if i choose 5 numbers from 1 to 50 what is the probability of at least one of the being between 41 and 50 (or any 10 numbers in the group)
thank you in advance and sorry if its very simple.

Do you mean 42,43,44,45,46,47,48,49?

Denis that is an excellent question. Although HT1992 tells us "i am not a mathematician" to a mathematician between means between.
X is between 41 & 50 means \(\displaystyle X\in\{42,43,44,45,46,47,48,49\}\).
On the other hand, if we want \(\displaystyle X\in\{41,42,43,44,45,46,47,48,49,50\}\) we say "X is from 41 to 50".

Now to answer HT1992 question \(\displaystyle 1-\dfrac{\binom{40}{5}}{\binom{50}{5}}\approx 0.689\) SEE HERE

Now I chose to use subsets: \(\displaystyle \dbinom{50}{5}\) is the number of ways to chose five ,any five, from fifty.

 

[HT1992]

Denis that is an excellent question. Although HT1992 tells us "i am not a mathematician" to a mathematician between means between.
X is between 41 & 50 means \(\displaystyle X\in\{42,43,44,45,46,47,48,49\}\).
On the other hand, if we want \(\displaystyle X\in\{41,42,43,44,45,46,47,48,49,50\}\) we say "X is from 41 to 50".

Now to answer HT1992 question \(\displaystyle 1-\dfrac{\binom{40}{5}}{\binom{50}{5}}\approx 0.689\) SEE HERE

Now I chose to use subsets: \(\displaystyle \dbinom{50}{5}\) is the number of ways to chose five ,any five, from fifty.

you are absolutely right i meant from 41 to 50 and i said the wrong word
your answer is right and i your number seems very consistent to my simulation ( i created a 1000 strings of 5 random numbers and counted the results and repeated for about 25 times the minimum i got was 651 out of the thousand and the most was 696 which makes some sense because the probability of getting lower than your number is higher than getting the higher number than yours because there are more numbers to change in the 689 numbers than in the remaining 311 numbers...am i right or wrong?)

 

[Denis]

My poor Denis, I worked with you for hours on how to count. Did you learn nothing? The list of number you wrote only contains 8 numbers while the OP wants 10 numbers.

....else yer a poor teacher...you're fired!!