Drawing numbers from an interval

Question

We draw $10$ numbers from an interval $[0,5]$. Compute probability, that at least two of these numbers will land in $[1,3]$.
So if $A-$ at least two numbers land in an interval $[1,3]$, then:
$A'-$ none of the numbers or only one will land in $[1,3]$, and it gives me:
$Bbb P(A')=frac{lambda_{10}([0,1)cup(3,5])+10 cdotlambda_{9}([0,1)cup(3,5])cdotlambda_1([1,3]) }{lambda_{10}([0,5])}$
where $lambda_a$ is a Lebesgue measure for a  cartesian product of intervals  in $Bbb R^a.$
Then
$Bbb P(A)=1-Bbb P(A')=1-frac{lambda_{10}([0,1)cup(3,5])+10 cdotlambda_{9}([0,1)cup(3,5])cdotlambda_1([1,3]) }{lambda_{10}([0,5])}=1-frac{3^{10}+10*3^9*2}{5^{10}}=0.9536$
Am I right? Thanks!

RobPratt · Answer

Yes, it's correct.  Here's another way to express it:
$$sum_{k=2}^n binom{n}{k}p^k (1-p)^{n-k}
=1-sum_{k=0}^1 binom{n}{k}p^k (1-p)^{n-k}
=1-(1-p)^n-n p (1-p)^{n-1}
$$
Now take $n=10$ and $p=(3-1)/(5-0)=2/5$.

Drawing numbers from an interval

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