You take the four Aces, four $2$'s, and four $3$'s from a standard deck of 52 cards, forming a set of $12$ cards. You then deal
all $12$ cards at random to four players, so that each player gets three cards. What is the probability that each player gets an Ace, a $2$, and a $3$?
1 answer:
Answer:
Probability each player gets an ace, a $2 and a $3 = 0.0374
Step-by-step explanation:
The total number of ways to divide the card in triples among four players = 369600 ways
The total number of ways to share the cards such that no card is repeated in each triple = 13824 ways
Probability each player gets an ace, a $2 and a $3 = 13824/369600
Probability each player gets an ace a $2 and a $3 = 0.0374
Note: Further explanation is provided in the attachment.
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Hi there!
The question sounds really complex, but the mathematical operations behind it is figuratively simple. Here, we are told to find 19 (18.7 rounded up is 19) percent of 1,000. To do this, we can set up a proportion to model the whole scenario.
Now, just solve for x by cross multiplying and simplifying.
Therefore, 190 people out of 1,000 people leave out pertinent information on their IRS forms. Hope this helped and have a terrific day!
The question sounds really complex, but the mathematical operations behind it is figuratively simple. Here, we are told to find 19 (18.7 rounded up is 19) percent of 1,000. To do this, we can set up a proportion to model the whole scenario.
Now, just solve for x by cross multiplying and simplifying.
Therefore, 190 people out of 1,000 people leave out pertinent information on their IRS forms. Hope this helped and have a terrific day!