B Why Does My Dice Probability Calculation Differ from Expected Results?

AI Thread Summary
The discussion centers on the calculation of the probability of rolling exactly one six with two dice. The initial calculation using the formula for non-mutually exclusive events led to an incorrect result due to double counting the outcome of rolling two sixes. Participants clarified that the correct approach involves recognizing the distinct cases for rolling one six, resulting in a probability of 10/36. The complement method was also suggested as a simpler alternative for calculating probabilities in such scenarios. Ultimately, the confusion stemmed from misinterpreting the event being calculated.
Thecla
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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?
 
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Seems like you are counting wrong.
 
You counted ##(6,6)## twice in ##P(A)+P(B)## but subtracted it only once.
 
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Thanks for your help. So it is 10/36=1/6+1/6-2/36
 
Thecla said:
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*}
 
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 said:
Seems like you are counting wrong.
Oops, sorry, I was wrong. I was counting “at least one 6” instead of “only one 6”
 
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