A five-card poker hand is dealt at random from a standard 52-card deck. Note the total number of possible hands is C(52,5)=2,598,960. Find the probabilities of the following scenarios: (a) What is the probability that the hand contains exactly one ace?

(a) Probability = 0.29947

Step-by-step explanation:

The probability of the hand containing exactly one ace would be:

Number of ways this can happen = 4C1 * 48C4      (using combinations)

Number of ways this can happen = 4 * 194580

Number of ways this can happen = 778,320

Total number possible hands = 2,598,960 (as stated in question)

Total probability of exactly one ace = Number of ways to get an ace / total number of ways

Total probability = 778320 / 2598960 = 0.29947

Thus, the probability of the hand containing exactly one ace will be 0.2994

Another way to solve this:

Probability of one ace and 5 other cards = = 0.059894

Number of ways to arrange 1 ace and 4 other cards = 5

Total probability = 0.0598 * 5 = 0.29947

she gotta workk that a 2

step-by-step explanation:

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)