What is the probability of getting a factor of 6 when you have 1 2 3 4 5 6

1 answer:

6 0

Hello,

I suppose you mean using all those digits one time.

The sum of the digits is a multiple of 3.
Every number formed with those digits is a multiple of 3.

To be divisible by 2 , the number must end with an even digit (2,4,6)

So probability is 3/6=1/2

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As suggested, we use the formula for exponential decay:

$A(t)=A_0\,e^{-k\,t}$

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$A(t)=A_0\,e^{-k\,t}\\\frac{A_0}{2} =A_0\,e^{-k\,*22}\\\frac{A_0}{A_0*2} =e^{-k\,*22}\\\frac{1}{2} =e^{-k\,*22}\\ln(\frac{1}{2})=-k\,*22\\ k=-\frac{ln(\frac{1}{2})}{22} \\k=0.0315$

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$A(t)=A_0\,e^{-0.0315\,t}\\0.9*A_0} =A_0\,e^{-0.0315\,t}\\\frac{0.9*A_0}{A_0} =e^{-0.0315\,t}\\0.9 =e^{-0.0315\,t}\\ln(0.9)=-0.0315\,t\\ t=-\frac{ln(0.9)}{0.0315} \\t=3.3447\,hours$