0
$\begingroup$

There are 35 kids who are given candy, the amount of candy each kid gets distributes $N(8,2^2)$. different amount of candy for each kid.

first question is, what's the probability that there are more than 300 candies overall?

I defined rv $s$ of $N(35*8,35^2*2^2)$ and $P(s>300)=1-P(s<=300)=1-Φ((300-35*8)/(2*35))=1-0.6123$

I hope that it's correct. Second part is where I'm stuck. out of 35 kids 15 are older and the others are younger. An adult marks the wraper of different candies from 5 different kids. What is the probability that all of the marked candies will belong to older kids?

I thought about approaching the problem as a binomial distribution of $Bin(5,15/35)$ but it doesn't seem correct beause not all kids have the same amount of candy. Would love to get some help as to how to solve this problem.

thank you.

$\endgroup$
2
  • 1
    $\begingroup$ The fact that the kids have different numbers of candies would, I think, be irrelevant. The way I read the problem, the adult selects $5$ distinct kids uniformly at random, without regard for the number of candies they have (well, we're tacitly assuming that they each have at least $1$, but that's not terribly important). $\endgroup$
    – lulu
    Commented Dec 8, 2020 at 19:34
  • 1
    $\begingroup$ Should say: I don't know what you mean by "different amount of candy for each kid." I am choosing to interpret that as saying that each child gets a random draw from the given distribution independently of all the other children. Of course, we expect many of the children to wind up with the same number. $\endgroup$
    – lulu
    Commented Dec 8, 2020 at 19:39

1 Answer 1

Reset to default
1
$\begingroup$

The probability to mark the candies exactly of 5 older kids is the probability to chose 5 older kids among the 35 total, say

$$\frac{\binom{15}{5}\binom{20}{0}}{\binom{35}{5}}$$

(an hypergeometric...)

The first point is wrong. The total candies distribution is the sum of 35 independent gaussian, thus the total amount of candies is

$$X=X_1+X_2+\dots+X_{35}$$

Assuming independence, you get

$$X\sim N(35\cdot8;4\cdot 35)$$

Thus

$$P(X>300)=1-\Phi\Bigg(\frac{300-35\cdot8}{\sqrt{35\cdot4}}\Bigg)\approx 4.55\%$$

$\endgroup$
6
  • $\begingroup$ even though that each kid has a different amount of candy I can still refer to them as identical when calculating the probability to choose 5 older kids? $\endgroup$
    – Mishe Mitasek
    Commented Dec 8, 2020 at 20:10
  • 1
    $\begingroup$ @MisheMitasek : yes, the question is only in choosing the kid. Note that first point is incorrect $\endgroup$
    – tommik
    Commented Dec 8, 2020 at 20:11
  • $\begingroup$ I see that you wrote 35 instead of $35^2$ that I wrote, but I see that the formula is $aX ~ N(aμ,a^2σ^2)$ so why is that? $\endgroup$
    – Mishe Mitasek
    Commented Dec 8, 2020 at 20:16
  • 1
    $\begingroup$ @MisheMitasek : you are dealing with $X=X_1+X_2+\dots+X_{35}$ and not with $X=35\cdot Y$. It is a different concept $\endgroup$
    – tommik
    Commented Dec 8, 2020 at 20:18
  • 1
    $\begingroup$ @MisheMitasek $X_1$ and $X_2$ have the same (identical) distribution but it doesn't mean that $X_1=X_2$. The total number of candies is the SUM of the candies each kid received...in a deterministic world you can say that this is $35\cdot X_1$ but in a random situation that is false. The sum of independent gaussians is still gaussian with mean sum of individual means ($35\mu$) and variance sum individual variances ($35\sigma^2$) $\endgroup$
    – tommik
    Commented Dec 8, 2020 at 20:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .