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Nimfa-mama [501]
2 years ago
5

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 $3$?

Mathematics
1 answer:
kap26 [50]2 years ago
8 0

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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A triangle weighs 3 grams and a circle weighs 6 grams. Find the weight of a square in Hanger A and the weight of a pentagon in H
Zanzabum

Answer:

Equation of Hanger A:

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Step-by-step explanation:

Note: This question is not complete, and lacks necessary data to answer this question. But for the sack of concept, we wil be solving this question using our own data.

Data given:

Triangle weight = 3 grams

Circle weight = 6 grams

There are two hangers A and B.

We have to find weight of the square in Hanger A and weight of the pentagon in Hanger B.

So, for your understanding I have drawn the two hangers A and B. In A we have triangle and circle on the LHS and the square on the right. On the other hand, we have pentagon on the right and triangle and circle.

Sketch is attached in the attachment.

Let's suppose: Hanger A is balanced, So,

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Circle weight = 6 grams = y

weight of the pentagon in Hanger B = p

weight of the square in Hanger A = z =?

So,

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z = 9 grams  = weight of the square in Hanger A

Similarly for Hanger B.

Equation for hanger B

x + y = p

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p = 9 grams

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