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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%

How to determine the probability?

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%

Read more about probability at:

https://brainly.com/question/25870256

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