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The digits 1,2,3,4,5,6,7 and 8 are arranged randomly to form a three-digit number find the probability that the number is even and greater than 800

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geerky42
So each digits can only be selected once.

Let's start counting desired outcome.

Hundred position need to be at least 8. There is only one possible digits. So we will go with 8 and then there are 7 remaining digits.

One position can only be 2,4,6. because we need number to be even. So that's 3 possible digits. We will take one. 6 digits remaining.

Ten position can be anything so that would be 6 possible digits.

So the number of desired outcomes is 3·6 = 18.

Now count all outcomes.

Hundred, ten, and one position can get any digits

So that's 8·7·6 = 336

So the probability would be 18 / 336 = 3 / 56 ≈ 0.0536

Hope this helps.
konrad509

[tex] |\Omega|=8\cdot7\cdot6=336\\
|A|=\underbrace{1\cdot6\cdot3}_{\text{8,any,even}}=18\\\\
P(A)=\dfrac{18}{336}=\dfrac{3}{56}=5\% [/tex]

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