Question

. 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

Answer 1

Answer:

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

$P=C_2^5\cdot 0.45^2\codt 0.55^3\\\\P=\frac{5!}{2!(5-2)!}\cdot 0.0337\\\\P=10\cdot 0.0337\\\\P=0.337$

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

$P=P(x\geq10)\\\\=1-P(x$

Threfore, the probability is P=0.77481.