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Random Probability and Statistics

1. Each vehicle in Mexico is either a truck, car or bicycle. Also each vehicle is red, green or yellow. We pick a vehicle at random. The following are known facts:

1. There's a 30% chance that vehicle is truck
2. There's 50% chance that vehicle is red
3. There's 20% chance that vehicle is a red truck
4. The vehicle is more likely to be a yellow truck than a green car

a. If we select a vehicle at random. What is the probability that it is red or a truck?
b. if we select a vehicle at random, what is the chance that it is neither red nor a truck?
c. if we select 100 vehicles, how many would be yellow or green?
d. let's say we select a vehicle at random. Show that the probability that the vehicle is either yellow or a bicycle is at least 0.4.

2. There are 75% guys & 25% girls in a class.

On a particular day, each guy has 40% chance of missing class & girl has 20 % chance of missing class.
a. what is the probability that a randomly chosen student is a girl and is not in class today?
B. what is the probability that a randomly chosen student will miss class today?
C. what is the probability that a randomly chosen student who is in class today is a girl?

3. How do you prove the following?

1. P (A+B+C) <= P (A) + P (B) + P(C) - 2P (ABC)
2. P (CD) <=P(C)

4. Consider an except that has n (finite) number of outcomes.

In how many ways can we choose n-1 of the 2^n events, so that the probabilities of all 2^n events can be computed from the n-1 known probabilities.

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Solution Preview

1. Suppose A represents truck, B represents red, then from condition 1,2,3, we have
P(A)=30%=0.3, P(B)=50%=0.5, P(AB)=20%=0.2
a. P(A+B)=P(A)+P(B)-P(AB)=0.3+0.5-0.2=0.6
b. P((not A) and (not B))=P(not(A+B))=1-P(A+B)=1-0.6=0.4
c. The probability for yellow or green is P(not B)=1-p(B)=1-0.5=0.5
So there will be 100*0.5=50 green or yellow vehicles.
d. Proof:
From b, there are 0.4 probability to be yellow or green cars or bicycles. So we consider yellow cars, yellow bicycles, green cars, green bicycles, denoted as C,D,E,F, repectively. So we have

Solution Summary

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