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3 years ago

What is the probability that a five-card poker hand contains exactly one ace?.

Mathematics

1 answer:

spin [16.1K]3 years ago

8 0

The probability of selecting exactly one ace is its likelihood

The probability that a five-card poker hand contains exactly one ace is 29.95%

<h3>How to determine the probability?</h3>

There are 4 aces in a standard deck of 52 cards.

The probability of selecting an ace would be:

p = 4/52

Also, there are 48 non-ace cards in the standard deck

So, the probability of selecting a non-ace after an ace has been selected is:

p = 48/51

The probability of selecting a non-ace up to the fifth selection are:

After two cards have been selected is: 47/50.
After three cards have been selected is: 46/49.
After four cards have been selected is: 45/48.

The required probability is then calculated as:

P(1 Ace) = n * (4/52) * (48/51) * (47/50) * (46/49) * (45/48)

Where n is the number of cards i.e. 5

So, we have:

P(1 Ace) = 5 * (4/52) * (48/51) * (47/50) * (46/49) * (45/48)

Evaluate

P(1 Ace) = 0.2995

Express as percentage

P(1 Ace) = 29.95%

Hence, the probability that a five-card poker hand contains exactly one ace is 29.95%

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Step-by-step explanation:

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If 5 cars arrive, 30 minutes to closing ; it takes 25 minutes to finish up ;

If 1 more car arrives, then it takes exactly 30 minutes

Hence, there will be no one in line at closing ;

IF 0 or 1 car arrives, if more than 1 car arrives, then there will be people in line at closing.

Hence, probability of waiting in line at closing :

1 - [p(0) + p(1)]

Using poisson :

P(x) = [(e^-μ) * (μ^x)] / x!

If x = 0

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P(0) = 0.0067379

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1 - 0.0404276

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