Respuesta :
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
[tex]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[/tex]
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
[tex]P=P(x\geq10)\\\\=1-P(x<10)\\\\=1-P(x\leq 9)\\\\=1-\sum_{x=0}^9 C_x^{25}\cdot 0.45^x\cdot 0.55^{25-x}\\\\=1-(0.00000032+0.0000066+0.00006+0.00041+0.00183+0.00629+0.03809+0.07013+0.10838)\\\\=1-0.22519\\\\\\=0.77481[/tex]
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:
- A) P (2) = (5! / 2! 3!) X (0.45) 2 x (0.55) 3 = 0.3332 x 100 = 33.32%
- B) P (10) = (25! / 10! 15!) X (0.45) 10 x (0.55) 15 = 0.8333 x 100 = 83.33%
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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