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?

Answers

Answer 1

Answer:

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: brainly.com/question/25870256

Answer 2

Answer:

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.

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Answers

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Answers

Answer:

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Answers

Answer:

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Answer:

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Answers

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The calculation can be done as follows

mass= 0.5 kg

time= 300 m/s

Kinetic energy= 1/2mv²

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Please see the link below for more information

brainly.com/question/10845930?referrer=searchResults

KE = 22500 J

KE= 1/2mv^2

1/2*.5*300^2

that's your answer AND formula.

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