Respuesta :
Answer:
a) 2.35 games.
b) Therefore, the probability is P=0.0229.
c) Therefore, the probability is P=0.71.
Step-by-step explanation:
We know that a roulette wheel has 38 slots, 18 are red, 18 are black, and 2 are green.You play five games and always bet on red.
Therefore, we have
[tex]p=\frac{18}{38}=0.47\\\\q=\frac{20}{38}=0.53\\[/tex]
a) We can you expect to win 5·0.47= 2.35 games.
b) We calculate the probability that you will win all five games.
[tex]P=C_5^5\cdot p^5\cdot q^0\\\\P=1\cdot 0.47^5 \cdot 0.53^0\\\\P=0.0229[/tex]
Therefore, the probability is P=0.0229.
c) We calculate the probability that you will win three or more games.
[tex]P=1-C_2^5\cdot 0.47^3\cdot 0.53^2\\\\P=1-10\cdot 0.47^3\cdot 0.53^2\\\\P=1-0.29\\\\P=0.71[/tex]
Therefore, the probability is P=0.71.
Using the binomial distribution, it is found that:
a) You expect to win 2.4 games.
b) 0.0239 = 2.39% probability that you will win all five games.
c) 0.5492 = 54.92% probability that you will win no more than two games.
For each game, there are only two possible outcomes. Either you win, or you lose. The probability of winning a game is independent of any other game, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- Five games, thus [tex]n = 5[/tex].
- 18 out of 38 spots are red, thus [tex]p = \frac{18}{38} = 0.4737[/tex].
Item a:
The expected value is:
[tex]E(X) = np = 5(0.4737) = 2.4[/tex]
You expect to win 2.4 games.
Item b:
This probability is P(X = 5), then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{5,5}.(0.4737)^{5}.(0.5263)^{0} = 0.0239[/tex]
0.0239 = 2.39% probability that you will win all five games.
Item c:
This probability is:
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
Then
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.4737)^{0}.(0.5263)^{5} = 0.0404[/tex]
[tex]P(X = 1) = C_{5,1}.(0.4737)^{1}.(0.5263)^{4} = 0.1817[/tex]
[tex]P(X = 2) = C_{5,2}.(0.4737)^{2}.(0.5263)^{3} = 0.3271[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0404 + 0.1817 + 0.3271 = 0.5492[/tex]
0.5492 = 54.92% probability that you will win no more than two games.
A similar problem is given at https://brainly.com/question/24863377
