When two unbiased dice are rolled one by one, what is the probability that either the first one is $2$ or the sum of the two is less than $5$?
a) $\dfrac 16$
b) $\dfrac 29$
c) $\dfrac 5{18}$
d) $\dfrac 13$
(Not homework, I'm doing some mock exams I found online)
Now I was pretty sure I got this. Take probability of rolling $2$ on the first one $\left(\frac 16\right)$ and the probability that the sum is $< 5 \left(\frac 6{36}\right)$, add them together and bam, $\frac 13$.
Wrong, seems the answer is c) $\frac5{18}$.
My problem is I've been trying to brush on my probability skills by googling and can see problems I made, such as the wording meaning that rolling a $2$ AND having a total of $5$ NOT being included in the result, but no matter what angle I attack from, I never end up with $\frac5{18}$.
So which little thing did I miss? Thanks very much for helping.
Best Answer
First consider things intuitively
We have the following table for adding the dice together
$$\begin{array}{|cc|cccccc|}\hline &&&&&D_1\\ &&1&2&3&4&5&6\\ \hline &1&\color{red}2&\color{red}3&\color{red}4&5&6&7\\ &2&\color{red}3&\color{red}4&5&6&7&8\\ D_2&3&\color{red}4&\color{red}5&6&7&8&9\\ &4&5&\color{red}6&7&8&9&10\\ &5&6&\color{red}7&8&9&10&11\\ &6&7&\color{red}8&9&10&11&12\\ \hline\end{array}$$
The red highlighted values are the ones which satisfy our conditions
We can count them to give us the answer $$P(D_1=2\text { OR }D_1+D_2<5)=\frac{10}{36}=\frac{5}{18}$$
Now consider things mathematically
We can count up $P(D_1=2)=\dfrac6{36}=\dfrac16$
$$\begin{array}{|cc|cccccc|}\hline &&&&&D_1\\ &&1&2&3&4&5&6\\ \hline &1&2&\color{red}3&4&5&6&7\\ &2&3&\color{red}4&5&6&7&8\\ D_2&3&4&\color{red}5&6&7&8&9\\ &4&5&\color{red}6&7&8&9&10\\ &5&6&\color{red}7&8&9&10&11\\ &6&7&\color{red}8&9&10&11&12\\ \hline\end{array}$$
We can also count up $P(D_1+D_2<5)=\dfrac 6{36}=\dfrac16$
$$\begin{array}{|cc|cccccc|}\hline &&&&&D_1\\ &&1&2&3&4&5&6\\ \hline &1&\color{red}2&\color{red}3&\color{red}4&5&6&7\\ &2&\color{red}3&\color{red}4&5&6&7&8\\ D_2&3&\color{red}4&5&6&7&8&9\\ &4&5&6&7&8&9&10\\ &5&6&7&8&9&10&11\\ &6&7&8&9&10&11&12\\ \hline\end{array}$$
However, note that while doing this, we counted two values twice, $(D_1,D_2)=(2,1)$ and $(D_1,D_2)=(2,2)$
$$\begin{array}{|cc|cccccc|}\hline &&&&&D_1\\ &&1&2&3&4&5&6\\ \hline &1&2&\color{red}3&4&5&6&7\\ &2&3&\color{red}4&5&6&7&8\\ D_2&3&4&5&6&7&8&9\\ &4&5&6&7&8&9&10\\ &5&6&7&8&9&10&11\\ &6&7&8&9&10&11&12\\ \hline\end{array}$$
So we must then subtract $\dfrac2{36}=\dfrac1{18}$ from our answer, giving
\begin{align}P(D_1=2\text { OR }D_1+D_2<5)&=P(D_1=2)+P(D_1+D_2<5)-P(D_1=2\text { AND }D_1+D_2<5)\\ &=\frac16+\frac16-\frac1{18}\\ &=\frac 5{18}\end{align}