Answer: a) 0.6561
b) 0.9999
Step-by-step explanation:
We use Binomial distribution here (Since each satellite is independent of each other.).
In Binomial distribution :[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex]
, where P(X) = probability of getting x successes in n trials.
p= probability of getting success in each trial.
As per given , we have
n= 4
p= 0.9
Let x = Number of satellites will correctly detect the next incoming missile.
a) The probability that all 4 satellites will correctly detect the next incoming missile:
[tex]P(x)=^4C_4(0.9)^4(1-0.9)^{0}[/tex]
[tex]P(x)=(1)(0.9)^4(1)=0.6561[/tex]
∴the probability that all 4 satellites will correctly detect the next incoming missile=0.6561
b) The probability that at least one of the four satellites will correctly detect the next incoming missile :
[tex]P(X\geq1)=1-P(X<1)\\\\=1-P(X=0)\\\\=1-^4C_0(0.9)^0(1-0.9)^4\\\\=1-(1)(1)(0.1)^4\\\\=1-0.0001\\\\=0.9999[/tex]
∴the probability that at least one of the four satellites will correctly detect the next incoming missile = 0.9999