Step 1
State the expression for the probability of an event to occur.
![\text{The probability of an event =}\frac{number\text{ of required events}}{\text{Total number of events}}](https://img.qammunity.org/2023/formulas/mathematics/college/46vbu5wonyk7ikqcpqijjuag485xre49lr.png)
Step 2
Find out the number of required events
Total number of events = 52 cards
![\begin{gathered} A\text{ deck of cards has} \\ 4\text{ Suites - }hearts,\text{ clubs, spade and diamond.} \\ \text{and each suite has} \\ \mleft\lbrace ace,\text{ 2,3,4,5,6,7,8,9, jack, qu}een,\text{ king}\mright\rbrace \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7gxpr4b2j7d6ssu0ecpb9hc4vz8t56e3v8.png)
Therefore, the number of aces = 4
The number of 6 = 4
Step 3
Find the probability of getting an ace or a 6.
![\begin{gathered} \text{The probability of getting an ace =}\frac{number\text{ of aces in a deck of cards}}{\text{Total number of cards}} \\ \text{The probability of getting an ace}=(4)/(52) \\ \text{The probability of g}etting\text{ a 6 =}\frac{number\text{ of 6 in a deck of cards}}{\text{Total number of cards}} \\ \text{The probability of g}etting\text{ a 6}=(4)/(52) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3vnnjj71h7f5wazogm6i0klmhbfudtqz7c.png)
Therefore,
![\text{The probability of drawing an ace or a 6 = }(4)/(52)+(4)/(52)=(8)/(52)=(2)/(13)](https://img.qammunity.org/2023/formulas/mathematics/college/8zkvmm0zaztq2t4n2wv16d94sgzqzeejdt.png)
Hence, Option D