Question

Suppose that two cards are dealt from a standard 52-card poker deck. let a be the event that the sum of the two cards is 8 (assume that aces have a numerical value of 1). how many outcomes are in a?

Answer 1

To find the number of outcomes in event A, where the sum of two cards is 8, consider all possible pairs of cards that add up to 8 and multiply by the number of suit combinations. There are three pairs (Ace and 7, 2 and 6, 3 and 5) and each pair has 16 suit combinations, totalling 48 outcomes.

Step-by-step explanation:

The student asked how many outcomes are in the event A, where A is the event that the sum of two cards dealt from a standard 52-card deck is 8. We assume that aces have a numerical value of 1. To determine the number of outcomes, we have to consider all pairs of cards that add up to 8:

• Ace (with a value of 1) and 7.
• 2 and 6.
• 3 and 5.

Each pair consists of cards from the four different suits (clubs, diamonds, hearts, spades), which can be combined in 4x4 ways (for each of the two cards). Thus, for each pair, there are 16 possible combinations (4 for the first card and 4 for the second card).

Therefore, the total number of outcomes in event A can be calculated as:

Number of outcomes = (Number of combinations for Ace and 7) + (Number of combinations for 2 and 6) + (Number of combinations for 3 and 5)

Number of outcomes = (16 combinations for Ace and 7) + (16 combinations for 2 and 6) + (16 combinations for 3 and 5)

Number of outcomes = 16 + 16 + 16

Number of outcomes = 48

Therefore, there are 48 outcomes in event A where the sum of two cards is 8.

Answer 2

The Combinations that equal 8 are: A+7, 2+6, 3+5, 4+4, 5+3, 6+2, 7+A

For instance there are 4 Aces and 4 sevens therefore = 4*4 = 16

So there are 16 possible combinations of A,7. Another 16 for 6,2 and another 16 for 3,5.

16*3=48

4C2=6 a special case for 4,4

Therefore:

48+6=54

a= 54