Econ 1005 Turorial Sheet 6

ECON 1005

WORKSHEET 6

1. A newspaper article claims that 44% of all airplanes arriving at Piarco Airport are late. Find the probability that, in a random sample of 7 airplanes:

i. Exactly two will arrive late

ii. At least 1 will arrive late

iii. At least 3 airplanes will arrive on schedule

i. P(exactly 2 will arrive late):

Using Binomial Formula: P(x)= nCxpxqn-x

Let x = no. of successes in n trials (no. of late arrivals)= 2

n = total number of trials = 7

p = prob. of success = 0.44

q = p – 1 = prob. of failure= 0.56

n – x = no. of failures in n trials = 7 – 2= 5

P( x=2)= 7C2(0.44)2(0.56)5

Px=2=7!2!(5!)(0.1936)(0.055)

=50402400.19360.055=0.22

ii. P(at least 1 will arrive late)= P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5) + P(x=6) + P(x=7)

P( x=1)= 7C1(0.44)1(0.56)6

Px=1=7!1!6!0.440.031=50407200.440.031=0.095

P( x=2)= 0.22 (from i.)

P( x=3)= 7C3(0.44)3(0.56)4

Px=3=7!3!4!0.0850.098=50401440.0850.098=0.292

P( x=4)= 7C4(0.44)3(0.56)4

Px=4=7!4!3!0.0370.176=50401440.0370.176=0.228

P( x=5)= 7C5(0.44)5(0.56)2

Px=5=7!5!2!0.0160.314=50402400.0160.314=0.106

P( x=6)= 7C6(0.44)6(0.56)1

Px=6=7!6!1!0.0070.56=50407200.0070.56=0.027

P( x=7)= 7C7(0.44)7(0.56)0

Px=7=7!7!0!0.0031=0.003

Therefore; P(at least one will arrive late)= 0.095 + 0.22 + 0.292 + 0.228 + 0.106 +

0.027 + 0.003 = 0.971 or 0.97

OR

P(at least 1 will arrive late) = 1 – P(x=0) = 1 – (7C0 (0.44) 0 (0.56) 7)= 1 – 0.01727

= 0.98

iii. P(at least 3 will be on time)= P(x=3) + P(=4) + P(x=5) + P(x=6) + P(x=7)

Let n = 7, p = 0.56 and q = 0.44

P( x=3)= 7C3(0.56)3(0.44)4

Px=3=7!3!4!0.1760.037=50401440.1760.037=0.228

P( x=4)= 7C4(0.56)3(0.44)4

Px=4=7!4!3!0.0980.085=50401440.0980.085=0.292

P( x=5)= 7C5(0.56)5(0.44)2

Px=5=7!5!2!0.0550.194=50402400.0550.194=0.224

P( x=6)= 7C6(0.56)6(0.44)1

Px=6=7!6!1!0.0310.44=50407200.0310.44=0.095

P( x=7)= 7C7(0.56)7(0.44)0

Px=7=7!7!0!0.0171=0.017

Therefore, P(at least...