From experience, an airline knows that only 76% of the passengers booked on a flight from New York to Los Angeles actually board their flight. A random sample of 10 booked passengers from New York to Los Angeles is chosen. Find the probability that 6 or 7 of them board their flight.Do not round your intermediate computations, and round your answer to three decimal places.

Para resolver este problema, utilizaremos la distribución binomial. La fórmula de la distribución binomial es:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

donde:

• $n$ es el número total de ensayos (en este caso, 10 pasajeros).
• $k$ es el número de éxitos deseados (en este

To solve this problem, we can use the binomial probability formula. The binomial probability formula is given by:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

where:

• $n$ is the number of trials (in this case, 10 passengers),
• $k$ is the number of successes (in this case, 6 or 7 passengers boarding),
• $p$ is the probability of success on a single trial (in this case, 0.76),
• $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$.

We need to find the probability that exactly 6 or 7 passengers board their flight.

First, calculate the probability for exactly 6 passengers boarding:

$P(X = 6) = \binom{10}{6} (0.76)^6 (0.24)^4$

Calculate the binomial coefficient:

$\binom{10}{6} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

Now, calculate the probability:

$P(X = 6) = 210 \times (0.76)^6 \times (0.24)^4$

$P(X = 6) = 210 \times 0.1934917632 \times 0.00331776$

$P(X = 6) = 210 \times 0.000641$

$P(X = 6) = 0.13461$

Next, calculate the probability for exactly 7 passengers boarding:

$P(X = 7) = \binom{10}{7} (0.76)^7 (0.24)^3$

Calculate the binomial coefficient:

$\binom{10}{7} = \frac{10!}{7!3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

Now, calculate the probability:

$P(X = 7) = 120 \times (0.76)^7 \times (0.24)^3$

$P(X = 7) = 120 \times 0.1479776 \times 0.013824$

$P(X = 7) = 120 \times 0.002046$

$P(X = 7) = 0.24552$

Finally, add the probabilities for 6 and 7 passengers boarding:

$P(6 \text{ or } 7) = P(X = 6) + P(X = 7)$

$P(6 \text{ or } 7) = 0.13461 + 0.24552$

$P(6 \text{ or } 7) = 0.38013$

Therefore, the probability that 6 or 7 of the 10 booked passengers board their flight is approximately 0.380, rounded to three decimal places.