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Triss [41]
3 years ago
14

You are dealt 5 cards from a standard deck of 52 playing cards. In how many ways can you get.....

Mathematics
1 answer:
tatiyna3 years ago
5 0

Answer:

Probability of Four of a Kind

The probability of being dealt four of a kind is 0.0002400960384. On average, four of a kind is dealt one time in every 4,165 deals.

Probability of Two Pair

On average, players get two pair about one time in every 21 deals.

Step-by-step explanation:

Probability of Four of a Kind

Let's execute the analytical plan described above to find the probability of four of a kind.

First, we count the number of five-card hands that can be dealt from a standard deck of 52 cards. This is a combination problem. The number of combinations is n! / r!(n - r)!. We have 52 cards in the deck so n = 52. And we want to arrange them in unordered groups of 5, so r = 5. Thus, the number of combinations is:

52C5 = 52! / 5!(52 - 5)! = 52! / 5!47! = 2,598,960

Hence, there are 2,598,960 distinct poker hands.

We count the number of ways that five cards can be dealt to produce four of a kind. It requires three independent choices to produce four of a kind:

Choose the rank of the card that appears four times in the hand. A playing card can have a rank of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace. For four of a kind, we choose 1 rank from a set of 13 ranks. The number of ways to do this is 13C1.

Choose one rank for the fifth card. There are 12 remaining ranks, from which we choose one. The number of ways to do this is 12C1.

Choose a suit for the fifth card. There are four suits, from which we choose one. The number of ways to do this is 4C1.

The number of ways to produce four of a kind (Num4) is equal to the product of the number of ways to make each independent choice. Therefore,

Num4 = 13C1 * 12C1 * 4C1 = 13 * 12 * 4 = 624

Conclusion: There are 624 different ways to deal a poker hand that can be classified as four of a kind.

Finally, we compute the probability. There are 2,598,960 unique poker hands. Of those, 624 are four of a kind. Therefore, the probability of being dealt four of a kind (P4) is:

P4 = 624 / 2,598,960 = 0.0002400960384

The probability of being dealt four of a kind is 0.0002400960384. On average, four of a kind is dealt one time in every 4,165 deals.

Probability of Two Pair

To find the probability for two pair, we execute the same analytical plan that we've used to compute the other probabilities.

There are 2,598,960 distinct poker hands.

Next, count the number of ways that five cards can be dealt to produce two pair. It requires five independent choices to produce two pair:

Choose the rank for cards of matching rank. A playing card can have a rank of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace. For two pair, we choose 2 ranks from a set of 13 ranks. The number of ways to do this is 13C2.

Choose the rank of the remaining non-matching card. There are 11 remaining ranks, from which we choose one. The number of ways to do this is 11C1.

Choose suits for the first two-card combination. There are four suits, from which we choose two. The number of ways to do this is 4C2.

Choose suits for the second two-card combination. There are four suits, from which we choose two. The number of ways to do this is 4C2.

Choose a suit for the non-matching card. There are four suits, from which we choose one. The number of ways to do this is 4C1.

The number of ways to produce two pair (Numtp) is equal to the product of the number of ways to make each independent choice. Therefore,

Numtp = 13C2 * 11C1 * 4C2 * 4C2 * 4C1

Numtp = 78 * 11 * 6 * 6 * 4 = 123,552

Finally, compute the probability. There are 2,598,960 unique poker hands. Of those, 123,552 are two pair. Therefore, the probability of being dealt two pair (Ptp) is:

Ptp = 123,552 / 2,598,960 = 0.04753901561

On average, players get two pair about one time in every 21 deals.

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PLEASE HELP! 20 POINTS 1) A ball is thrown starting at a time of 0 and a height of 2 meters. The height of the ball follows the
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1) The height of the ball from 0 to 5  seconds are;

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2)  The correct option is;

D. -16·t² + 25·t + 1

Step-by-step explanation:

1) The equation of motion of the ball is given as follows;

H(t) = -4.9·t² + 25·t + 2

The height of the ball from 0 to 5 seconds are;

H(0) = -4.9×(0)² + 25×(0) + 2 = 2

H(1) = -4.9×(1)² + 25×(1) + 2 = 22.1

H(2) = -4.9×(2)² + 25×(2) + 2 = 32.4

H(3) = -4.9×(3)² + 25×(3) + 2 = 32.9

H(4) = -4.9×(4)² + 25×(4) + 2 = 23.6

H(5) = -4.9×(5)² + 25×(5) + 2 = 4.5

Therefore, we have;

The height of the ball are

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2) Given that the equation of the ball is that of a projectile motion, such as follows;

h = h₀ + v₀·sin(θ₀)·t - 1/2·g·t² which is equivalent to h = -1/2·g·t²+ h₀+v₀·sin(θ₀)·t

it is best represented by the quadratic equation of an upside down parabola which is option D. -16·t² + 25·t + 1

6 0
3 years ago
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EastWind [94]

Answer:

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Step-by-step explanation:

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<u>Add both together</u>

128/3 + 512 ≈ 646.04cm³

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