. In the game of blackjack as played in casinos in Las Vegas, Atlantic City, Niagara Falls, as well as many other cities, the dealer has the advantage. Most players do not play very well. As a result, the probability that the average player wins a hand is about 45%. Assume each hand is independent. Find the probability that the average player wins a. twice in 5 hands b. ten or more times in 25 hands

a) Threfore, the probability is P=0.337.

b) Threfore, the probability is P=0.77481.

Step-by-step explanation:

We know that the probability that the average player wins a hand is about 45%. Assume each hand is independent.

a) We calculate the  probability that the average player wins twice in 5 hands.

We know that p=0.45 ⇒ q=1-0.45=0.55

Threfore, the probability is P=0.337.

b) We calculate the  probability that the average player wins ten or more times in 25 hands. We get

Threfore, the probability is P=0.77481.

The probability that the average player wins twice in 5 hands is 33.32%, while the probability that the average player wins ten or more times in 25 hands is 83.33%.

Since in the game of blackjack the probability that the average player wins a hand is about 45%, to find the probability that the average player wins A) twice in 5 hands, and B) ten or more times in 25 hands, you must perform the following calculation, applying the binomial distribution theory:

Therefore, the probability that the average player wins twice in 5 hands is 33.32%, while the probability that the average player wins ten or more times in 25 hands is 83.33%.

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answer: the answer is 35%

i believe that the answer is

b. y-1=-2(x-4)

hope this !