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When playing the card game Blackjack, multiple decks are used and reshuffled often so that the outcomes of the cards dealt are approximately independent. When a player receives two cards that are a combination of an ace and a face card, this is called a “natural blackjack” and automatically wins. A natural blackjack should occur in 4.5% of the rounds played. What is the probability that a player plays 20 rounds of Blackjack and gets two or more natural blackjacks?

Respuesta :

Answer:

The correct answer is (C) or .227. The number of natural blackjacks (X) follows a binomial distribution with n=20, p=0.045.

Explanation:

P(X≥2)=1−P(X≤1)=1−0.773=0.227.

Lanuel

The probability that a player plays 20 rounds of Blackjack and gets two or more natural blackjacks is 0.227.

Given the following data:

How to calculate the probability.

In this scenario, we can deduce that the number of natural Blackjacks (B) is in tandem with a binomial distribution with the following parameters:

  • Sample mean, n = 20.
  • Percentage, P = 4.5% = 0.045.

Mathematically, the probability that a player plays 20 rounds of Blackjack and gets two or more natural blackjacks is given by:

P(B ≥ 2) = 1 - P(B ≤ 1)

P(B ≥ 2) = 1- 0.773

P(B ≥ 2) = 0.227.

Read more on probability here: https://brainly.com/question/25870256

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