Recall that the numbers in a die are 1,2,3,4,5,6.
![S=\mleft\lbrace1,2,3,4,5,6\mright\rbrace](https://img.qammunity.org/2023/formulas/mathematics/college/ta9o8eud34ylebxl53f500644j2qr7if3p.png)
Hence the number of possible outcomes is 6.
![n(S)=6](https://img.qammunity.org/2023/formulas/mathematics/college/l0ahxeiwa2tasbzib8tmh1gr5em0elbc4r.png)
We need a number less than 3. Let A be this event.
![A=\mleft\lbrace1,2\mright\rbrace](https://img.qammunity.org/2023/formulas/mathematics/college/akd9tuw002sg4xd9cmdmtctfd26qbuc4fp.png)
The favorable outcome is 2.
![n(A)=\mleft\lbrace1,2\mright\rbrace](https://img.qammunity.org/2023/formulas/mathematics/college/iodvxojjajfy2l5lxjjyxhplnyqb7kk3sm.png)
Since there are 1,2 less than 3 in a die.
![P(A)=\frac{Favourable\text{ outcomes}}{\text{Total outcomes}}=(n(A))/(n(S))](https://img.qammunity.org/2023/formulas/mathematics/college/yw0nnh5otn6134afd9y3dp8t3atu92l3uq.png)
Substitute n(A)=2 and n(S)=6, we get
![P(A)=(2)/(6)=(1)/(3)=0.333](https://img.qammunity.org/2023/formulas/mathematics/college/kbn0nizkojrvo2gq7gbuolrtmukadzk90n.png)
Hence the required probability is 0.333.