Question

A standard number cube is tossed. Find the probability of getting P(greater than 2 or less than 4).1/215/66

Answer 1

The probability of getting P(greater than 2 or less than 4) is given by

$\begin{gathered} P(\text{greater than 2 OR less than 4)=P(greater than 2)}+P(\text{less than 4)-P(greater than 2 AND less than 4)} \\ \text{which is equal to} \\ P(\text{greater than 2 OR less than 4)=P(greater than 2)}+P(\text{less than 4)-P(3)} \end{gathered}$

because the a number greater than 2 and less than 4 is 3.

Since the cube has 6 faces, the probability of getting one face is 1/6, then we have

$\begin{gathered} \text{P(greater than 2)}=P(3)+P(4)+P(5)+P(6) \\ \text{P(greater than 2)}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6} \\ \text{P(greater than 2)}=\frac{4}{6} \end{gathered}$

Similarly,

$\begin{gathered} P(\text{less than 4)}=P(3)+P(2)+P(1) \\ P(\text{less than 4)}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6} \\ P(\text{less than 4)}=\frac{3}{6} \end{gathered}$

since P(3)= 1/6, we have

$\begin{gathered} P(\text{greater than 2 OR less than 4)=P(greater than 2)}+P(\text{less than 4)-P(3)} \\ P(\text{greater than 2 OR less than 4)=}\frac{4}{6}+\frac{3}{6}-\frac{1}{6} \end{gathered}$

which gives

$P(\text{greater than 2 OR less than 4)=}\frac{7-1}{6}=\frac{6}{6}=1$

Therefore, the answer is 1.