createsure wrote:
1) 13/100
Yes that's correct. It's a pretty trivial question!
Math Probability Questions :)
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Jul 29, 2020
0
#3
LOL, just tried deleting to give others a chance, yes it was easy! Thanks.
Jul 29, 2020
0
#4
5) 21600 out of 46656 (a hair under 46.3%)
(solved non-electronically, except to convert to a percentage)
2) won’t the probability of a^2 + b^2 =< 1 be zero? Both a and b are either 1 or bigger than 1. I don’t know how the y and x inequality is related to a and b though😅. Can anyone help? 
Jul 29, 2020
0
#6
Minecraft710 wrote:
2) won’t the probability of a^2 + b^2 =< 1 be zero? Both a and b are either 1 or bigger than 1. I don’t know how the y and x inequality is related to a and b though😅. Can anyone help?
no, I had the exact same thought... the first part is zero, and WTH is the second part even talking about? LOL. It's like part of the problem is missing or something.
Jul 31, 2020
0
#7
1) 7/50 (4,11,18,25,32,39,46,53,60,67,74,81,88,95)
if u divide 7 into 4 the remainder is 4
The answer of (2) should be (pi - 2)/400
Aug 9, 2020
0
#9
3) The answer is zero probability. Either that or I don't understand the wording. I take it to mean that NONE of the kids have 2, 1, or zero candies. In other words, everyone must have at least 3. Which is not possible, hence zero probability.
Aug 10, 2020
0
#12
This_Is_Amazing wrote:
a possible configuration can be 1,1,1,8
What does the phrase "at most" mean then? The first kid has 1 candy in your example. Does he have "at most 2 candies"? Yes, he does.
So, we can pretend that the question says "none of them have 2 candies" which would be a fairly easy problem to work out. But that is NOT what it says. Just as the phrase "at least" means that number or more, the phrase "at most" means that number or less. And 1 is less than 2, so your example is invalid.
4, 0
1) Let x be a random positive integer between 1 and 100 inclusive. Find the probability x leaves a remainder of 4 when divided by 7.
(SOLVED)
2) Let a and b be random reals between 0 and 10 inclusive. Find the probability that a^2+b^2 =< 1 and y >= 1 - x.
3) 11 indistinguishable candies are randomly distributed to 4 kids. Find the probability that none of the kids have at most 2 candies.
4) Let a, b, and c be reals between 10 and -10 inclusive. Find the probability a^2+b^2+c^2 =< 1 and exactly one of the variables a, b and c are positive.
5) There are 6 dice numbered with the integers between 0 and 5 inclusive that are fair. Find the probability that when all of the dice are rolled simultaneously, the sum of the numbers shown on the dice is 15. (Its okay to express your answer as binomials, if you do happen to do so, use the form binomial(a, b) where binomial(a, b) defines the number of ways to choose a committee of b people from a people.)
Thank you,
SuperJupiter5 (Edited username)