One card is drawn at random from a standard pack of 52 playing cards. What is the probability that the card drawn is neither a spade nor a Jack?

Introduction / Context:
This problem checks basic set based probability with playing cards. Instead of asking for one particular suit or rank, it asks for cards that are not spades and also not Jacks. To solve this, we use the idea of excluding unwanted cards from the total and then computing the probability for the remaining cards.

Given Data / Assumptions:

• A standard deck contains 52 cards.
• There are 4 suits, so there are 13 spades.
• There are 4 Jacks in the deck, one from each suit.
• The Jack of spades is counted both as a spade and as a Jack.
• One card is drawn at random and all cards are equally likely.
• We want the card to be neither a spade nor a Jack.

Concept / Approach:
We use the complement principle through set operations. First we count how many cards are in the union of the set of spades and the set of Jacks. These are the cards that we do not want. Then we subtract this count from the total number of cards to get the allowed cards. Finally, the probability is the number of allowed cards divided by 52.

Step-by-Step Solution:
Number of spades = 13. Number of Jacks = 4. Jack of spades is common to both sets, so it is counted twice above. Cards that are spades or Jacks = 13 + 4 - 1 = 16. Cards that are neither spades nor Jacks = 52 - 16 = 36. Required probability = 36 / 52. Simplify 36 / 52 by dividing numerator and denominator by 4 to get 9 / 13.

Verification / Alternative check:
We can also think directly about suits and ranks. Non spade suits are hearts, diamonds and clubs, which together have 39 cards. From these, we must remove the remaining 3 Jacks: Jack of hearts, Jack of diamonds and Jack of clubs. So allowed cards = 39 minus 3 = 36. Dividing by 52 again gives the same probability 9 / 13, which verifies the previous calculation.

Why Other Options Are Wrong:
4/13 corresponds to 16 favourable cards, which actually equals the number that are spades or Jacks, not those excluded from that union. 10/13 and 8/13 are larger than 9/13 and would require more allowed cards than the 36 that really remain after removing all spades and all Jacks.

Common Pitfalls:
A common mistake is to subtract 13 and 4 directly from 52 without correcting for the double counted Jack of spades, which gives 52 minus 17 instead of 52 minus 16. Another error is to treat spades and Jacks as independent, which they are not, since suits and ranks combine in specific cards. Always remember to adjust for overlap when applying the union formula for sets.

Final Answer:
The probability that the card drawn is neither a spade nor a Jack is 9/13.