Respuesta :
Answer:
[tex](1 - \sum_{j = 0}^{k-1} {n \choose j} * {p_1}^j * (1-p_1)^{n-j}) * a + (1 - \sum_{j=0}^{k-1} {n \choose j} * {p_2}^j * (1-p_2)^{n-j}) * (1-a)[/tex]
Step-by-step explanation:
We will call C the number of components that will function tomorrow. R is the event 'tomorrow rains', and D is the event 'tomorrow is a dry day'. The satellite system will work if [tex] C \geq k [/tex].
We can calculate the probability of the event [tex] \textbf{textit{C \geq k}} [/tex] dividing in 2 cases, depending on if it rains tomorrow (R), of if is a dry day (D). According to the theorem of total probability, we have this formula
[tex] P(C \gek k) = P(C \geq k | R) * P(R) + P(C \geq k | D) * P(D) [/tex]
Since D is the complementary event of R, we have [tex] P(D) = 1- P(R) = 1-a[/tex]
Note that if we assume that it will rain (or not) tomorrow, then C is a random variable of binomial distribution with parameters n and [tex] p_1 [/tex] (or [tex] p_2 [/tex]). To calculate the probability of C being greater than k, we can take probability of the complementary event, in other words, the probability of the event 'C is any j with j between 0 and k-1' and substract it from 1. Therefore
- [tex] P(C \geq k | R) = 1 - \sum_{j = 0}^{k-1} {n \choose j} * {p_1}^j * (1-p_1)^{n-j} [/tex]
- [tex] P(C \geq k |D) = 1 - \sum_{j=0}^{k-1} {n \choose j} * {p_2}^j * (1-p_2)^{n-j} [/tex]
Using this calculations, we can conclude that
[tex]P(C \geq k) = P(C \geq k |R) * P(R) + P(C \geq k |D) * P(D) \\ = (1 - \sum_{j = 0}^{k-1} {n \choose j} * {p_1}^j * (1-p_1)^{n-j}) * a + (1 - \sum_{j=0}^{k-1} {n \choose j} * {p_2}^j * (1-p_2)^{n-j}) * (1-a)[/tex]
I hope it helps you!
