Probability in a roll of dice

[Thecla]

TL;DR Summary

What is probability of throwing one six in a throw of pair of dice?

I was trying to calculate the probability of throwing only one six when throwing a pair of dice.
Using the formula for non-mutually exclusive events : P(A)+P(B)-P(A,B) I get 1/6+1/6-1/36=11/36
but when I count all the 36 possibilities on paper I get 10/36 ways of getting only one 6. What am I doing wrong?

 

[Dale]

Seems like you are counting wrong.

 

[fresh_42]

You counted $(6,6)$ twice in $P(A)+P(B)$ but subtracted it only once.

 

[Thecla]

Thanks for your help. So it is 10/36=1/6+1/6-2/36

 

[fresh_42]

Thanks for your help. So it is 10/36=1/6+1/6-2/36

Yes. A common trick for those questions is to use the complement statement. We have $5^2$ cases without a $6$ plus $(6,6)$, which leaves $10$ positive cases.
\begin{align*}
P((\lnot A \wedge \lnot B)\vee (A\wedge B) )&=P(\lnot A \wedge \lnot B)+P(A\wedge B)\\&=P(\lnot A)\cdot P(\lnot B)+P(A)\cdot P(B)\\&=\dfrac{5}{6}\cdot\dfrac{5}{6}+\dfrac{1}{6}\cdot\dfrac{1}{6}\\&=\dfrac{26}{36}
\end{align*}

 

[PeroK]

Simplest of all is $$p = \frac 1 6 \cdot \frac 5 6 + \frac 5 6 \cdot \frac 1 6 = \frac {10}{36}$$based on whether a six is thrown on the first die or not.

 

[Dale]

Seems like you are counting wrong.

Oops, sorry, I was wrong. I was counting “at least one 6” instead of “only one 6”