Question

If two cards are drawn at random simultaneously from a well shuffled pack of 52 playing cards, then the probability of getting a card having a composite number and a card having a number which is a multiple of 3 is:

• A. \( \frac{94}{663} \)
• B. \( \frac{62}{663} \)
• C. \( \frac{102}{663} \)
• D. \( \frac{64}{663} \) \bigskip

Hint:

When solving probability problems with a deck of cards, first determine the total number of outcomes and then count the favorable cases according to the given criteria, such as composite numbers or multiples of a certain value.

Answer 1

The Correct Option is C

We need to determine the probability of drawing two cards where: - One card has a composite number.
- One card has a multiple of 3.
Step 1: Identify Composite and Multiple of 3 Cards A standard deck has numbers from 2 to 10 (excluding face cards) in each suit (hearts, diamonds, spades, clubs). - Composite numbers: \( 4, 6, 8, 9, 10 \)
- Multiples of 3: \( 3, 6, 9 \)
Each number appears in 4 suits, so: - Cards with composite numbers: \[ 4(5) = 20 \text{ cards} \] - Cards with multiples of 3: \[ 4(3) = 12 \text{ cards} \] - Overlap (cards that are both composite and multiples of 3, i.e., 6 and 9): \[ 4(2) = 8 \text{ cards} \] Using set notation: \[ |C| = 20, \quad |M| = 12, \quad |C \cap M| = 8 \] Using the formula for choosing one from each category: \[ \text{Ways to choose (Composite, Multiple of 3)} = |C| \cdot |M| - |C \cap M| \cdot (|C \cap M| - 1)/2 \] \[ = (20 \times 12) - \binom{8}{2} \] \[ = 240 - \frac{8(7)}{2} \] \[ = 240 - 28 = 212 \]

Step 2: Compute Probability Total ways to choose 2 cards from 52: \[ \binom{52}{2} = \frac{52 \times 51}{2} = 1326 \] \[ P = \frac{102}{663} \]