At a car park there are 100 vehicles, 60 of which are cars, 30 are vans and the remaining are lorries. If every vehicle is equally likely to leave, find the probability of car leaving second if either a lorry or van had left first:
Answers
Let S be the sample space and A be the event of a van leaving first. n(S) = 100 n (A) = 30
Probability of a van leaving first: P (A) = 30/100 = 3/10
Let B be the event of a lorry leaving first. n (B) = 100 - 60 - 30 = 10 Probability of a lorry leaving first: P (B) = 10/100 = 1/10
If either a lorry or van had left first, then there would be 99 vehicles remaining, 60 of which are cars.
Let T be the sample space and C be the event of a car leaving. n (T) = 99 n(C) = 60
Probability of a car leaving after a lorry or van has left: P(C) = 60/99 = 20/33
Given : At a parking, there are 100 vehicles, 60 of which are cars, 30 are busses and the remainder lorries. every vehicle is equally likely to leave,
To Find : the probability of car leaving second if either a lorry or van had left first:
Solution:
Total vehicles = 100
Cars = 60
Buses = 30
Trucks = 100 - 60 - 30 = 10
Not cars = 100 - 60 = 40
car leaving second = car leaving not 1st x car leaving 2nd
= (40/100) * (60/99)
= 24/99
=8/33
= 0.2424
0.2424 = 8/33 is the probability of car leaving second if either a lorry or van had left first:
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