help a retard

[Anonymous No.6139761]

simple question:

If I decide to make 10 posts on 4chan, what is the chance that at least one of those 10 posts will end with a zero?

[Anonymous No.6139767]

Assuming it isn't on /b/ where dubs are removed, the answer is 1/10.

[Anonymous No.6139772]

simple question:

If I decide to make 10 posts on 4chan, what is the chance that at least one of those 10 posts will be a shitpost?

[Anonymous No.6139771]

>>6139767
can you explain what formula you use to get this?

suppose I switched the number of posts to 11. how do I calculate the probability that one of them will be 0?

[Anonymous No.6139773]

>>6139767
Maybe I'm over thinking this, but if each post has 7 numbers, and in this case we will count zero as a number since it is acting as a place holder, you actually have 11 different numbers that your post could end in, you've made ten posts. So multiply all these things together and you have a 1 in 770 chain that your post will end in zero. Or a .001% chance. I'm probably wrong and look like an idiot.

[Anonymous No.6139775]

>>6139761
1- (10C9) * 0.1^1 * 0.9^9 = 61.3%

[Anonymous No.6139790]

>>6139773

not sure if trolling or retarded...

[Anonymous No.6139792]

>>6139790
I'm not trolling, just retarded.

[Anonymous No.6139801]

>>6139771
>can you explain what formula you use to get this?

How does it feel to suffer from downs?

[Anonymous No.6139807]

>>6139761
>what is binomial distribution

[Anonymous No.6139810]

>>6139801
I don't have downs, I am just ignorant of probability theory.

And it seems curiosity is slowly starting to eat me alive-

[Anonymous No.6139817]

>>6139775
>at least one

[Anonymous No.6139819]

>>6139761
summation of x from 1 to 10 ((0.1^x)(0.9^(10-x))(10Cx))

[Anonymous No.6139825]

>>6139817
1- (10Choose0) * 0.1^0 * 0.9^10 = 65%

[Anonymous No.6139863]

chance that at least one post will end with 0
= 1 - (chance that none of the posts end in 0)
= 1 - (.9)^10
~ .651322

what
>>6139825
said

[Anonymous No.6139887]

>>6139863
>>6139825

Thanks!

[Anonymous No.6139903]

(1/10)^10 = 1/10000000000 chance to not get atleast 1 ending in 0

[Anonymous No.6139904]

>>6139903
in other words youre more likely to die in a car crash than not get a 0 in 10 rolls.

[Anonymous No.6139920]

>>6139903
>>6139904

No. That would be the probability that they ALL end in 0.

Also, this question is already answered correctly more than once.

[Anonymous No.6139925]

>>6139903
>>6139904
Probability of not getting 0 on any given post=0.9

Probability of not getting 0 on 10 consecutive posts=0.9^10=0.349

Probability of not getting 0 on 10 consecutive posts=0.9^10=0.349

Probability of getting 0 on any given post=0.1

Probability of getting 0 on 10 consecutive posts=0.1^10=1e-10

("0 on" refers to 0 as the last digit of any given post).