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Answer:
P(at least 4) = 0.74431
Step-by-step explanation:
Here
Probability of the flights on time is = 70% = 70/100= 0.7
q= 0.3
n= 6
As the probability of success and failure is constant and the n is also fixed then the binomail probability distribution can be used to solve this.
P (x= 4) = 6c4 *(0.7)^4 (0.3)^2= 15*0.2401* 0.09= 0.324135
P (x= 5) = 6c5 *(0.7)^5(0.3)^1= 6*0.16807* 0.3=0.302526
P (x= 6) = 6c6 *(0.7)^6 (0.3)^0= 0.117649P
P (x= 0) = 6c0 *(0.7)^0 (0.3)^6= 0.000729
P (x= 1) = 6c1 *(0.7)^1 (0.3)^5=0.010206
P (x= 2) = 6c 2*(0.7)^2 (0.3)^4=0.059535
P (x= 3) = 6c3 *(0.7)^3 (0.3)^3=0.18522
Probability of at least 4 means that probability of greater than 4 inclusive of 4.
This can be obtained in two ways.
First is by subtracting the probabilities of less than 4.
Second is by adding probabilities greater than 4 including 4.
We get the same answer with any of the two methods used.
P(at least 4) = 1- P(x<4)
= 1- [ 0.000729+ 0.010206 +0.059535+ 0.18522]
= 1- 0.25569
= 0.74431
P (at least 4) = P(x=4) +P(X=5) + P(x=6)
=0.324135+ 0.302526+ 0.117649
=0.74431