Question

perform the following complete probability exercise on a calculator Banco del País recently started a new credit program. Customers who meet certain credit requirements can obtain a credit card that is accepted by area merchants. Previous records indicate that 25% of all applicants for this type of card are rejected. If a total of 5 applications are received in one day, what is the probability that exactly 3 will be rejected? Select one: to The probability that exactly 3 are rejected is 98.44% b. The probability that exactly 3 are rejected is 10.35% c. The probability that exactly 3 are rejected is 0.09% d. The probability that exactly 3 are rejected is 8.79%

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Answer to a math question perform the following complete probability exercise on a calculator Banco del País recently started a new credit program. Customers who meet certain credit requirements can obtain a credit card that is accepted by area merchants. Previous records indicate that 25% of all applicants for this type of card are rejected. If a total of 5 applications are received in one day, what is the probability that exactly 3 will be rejected? Select one: to The probability that exactly 3 are rejected is 98.44% b. The probability that exactly 3 are rejected is 10.35% c. The probability that exactly 3 are rejected is 0.09% d. The probability that exactly 3 are rejected is 8.79%

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Brice
4.8
113 Answers
La probabilidad de que una solicitud sea rechazada es del 25%, por lo tanto, la probabilidad de que una solicitud sea aceptada es del 75%.

Usaremos la distribución binomial para calcular la probabilidad de que exactamente 3 de las 5 solicitudes sean rechazadas.

La fórmula para la distribución binomial es:

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

Donde:
- n = 5 (número total de solicitudes)
- k = 3 (número de solicitudes rechazadas)
- p = 0.25 (probabilidad de ser rechazada)
- 1-p = 0.75 (probabilidad de ser aceptada)

Sustituyendo estos valores en la fórmula:

P(X = 3) = \binom{5}{3} \cdot (0.25)^3 \cdot (0.75)^{2}

Calculamos \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \cdot 4}{2 \cdot 1} = 10

Entonces,

P(X = 3) = 10 \cdot (0.25)^3 \cdot (0.75)^2 = 10 \cdot 0.015625 \cdot 0.5625 = 0.08789

Por lo tanto, la probabilidad de que exactamente 3 de las 5 solicitudes sean rechazadas es del 8.79%.

\boxed{\text{La probabilidad de que exactamente 3 sean rechazadas es de un 8,79%}}

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