Answer: 0.9887
Step-by-step explanation:
Let p be the probability that flights are on time.
As per given, p =70% = 0.7
Sample size : n = 8
Let x be a binomial variable that represents the number of fights are on time ( i.e. success).
Using binomial formula: [tex]P(X=x) = \ ^nC_xp^x(1-p)^{n-x}[/tex]
Now, the probability that at least 3 flights are on time = [tex]P(X\geq 3)=1-P(X<3)\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\\\\=1-[^8C_0(0.7)^0(0.3)^8+^8C_1(0.7)^1(0.3)^7+^8C_2(0.7)^2(0.3)^6]\\\\=1-[(0.3)^8+8(0.7)(0.3)^7+\dfrac{8!}{2!6!}(0.7)^2(0.3)^6]\\\\=1-[0.00006561+0.00122472+0.01000188]\\\\=1-0.01129221\\\\=0.98870779\approx\ 0.9887[/tex]
The probability that at least 3 flights are on time = 0.9887
