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)}=(1)/(6)+(1)/(6)+(1)/(6)+(1)/(6) \\ \text{P(greater than 2)}=(4)/(6) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8ki5dswk31wjp9tlqv1a126gldw4ylfvae.png)
Similarly,
![\begin{gathered} P(\text{less than 4)}=P(3)+P(2)+P(1) \\ P(\text{less than 4)}=(1)/(6)+(1)/(6)+(1)/(6) \\ P(\text{less than 4)}=(3)/(6) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cgoa21ns2pqx7wjs7acsllxxkbpvn0t6gi.png)
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)=}(4)/(6)+(3)/(6)-(1)/(6) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6c0uc0phth2w3pq5mudtxpyn9ro9sipib5.png)
which gives
![P(\text{greater than 2 OR less than 4)=}(7-1)/(6)=(6)/(6)=1](https://img.qammunity.org/2023/formulas/mathematics/college/k4mcr90gdf16jrxha7h5pxap2j7adnaj6f.png)
Therefore, the answer is 1.