Anonymous
Anonymous asked in Science & MathematicsMathematics · 4 weeks ago

# math hw probability please someone ?

Using the non-standard deck from the previous problem, suppose you play a version of poker where you draw a random hand of three cards (without replacement). In determining the hand, the order of the cards does not matter.

a.) What is the probability of getting a triple-nova , meaning that all the cards are 8s or all the cards are 9s? Round to three decimal places.

b.) What is the probability of getting a three-of-a-kind, meaning three cards of the same denomination? Round to three decimal places.

For c.) and d.), please round to two decimal places.

c.) What is the probability of getting a pair, meaning 2 cards of the same denomination (different colors) and any other card of a different denomination?

d.) What is the probability of getting a diversity, meaning that all three cards are of different denominations?

(help with any of these)

Relevance
• Read the problem and tell me what's wrong

(hint: "previous problem")

So I'm going to assume this deck of yours has no JQK or jokers (40 cards)

a.) What is the probability that all the cards are 8s or all the cards are 9s?

There are 4x 8s and 4x 9s. From those 8 cards, we want 3, and order doesn't matter so we can use "choose"

8C3 / 40C3

b.) What is the probability of getting a three-of-a-kind?

From any denomination (n=10), we want 3 cards.

(10C1 * 4C3) / 40C3

c.) What is the probability of getting a pair (different colours) and any other card of a different denomination?

From any denomination (n=10), we want 2 cards of different colours. Say we chose '5'. There are 2 black cards and 2 red cards. There are 2C1*2C1 = 4 ways of getting different coloured pairs. Then we want some random card of another denomination, so just choose any of the remaining 9 (9C1) and take a card (4C1)

( 10C1 * 2C1 * 2C1 * 9C1 * 4C1 ) / 40C3

d.) What is the probability of getting all three cards of different denominations?

(10C1 * 4C1 * 9C1 * 4C1 * 8C1 * 4C1) / 40C3

This may be too late to tell you, but I'm terrible at probability and likely got all the above wrong