Probability - Advanced Engineering Math - Tutorial Slides, Slides of Engineering Mathematics

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Probability

A box contains 10 screws, three of which are defective. Two screws are drawn at random. Consider both sampling with and without replacement, find the probability that neither of the two screws is defective.

Consider events: A. First drawn screw nondefective; B. Second drawn screw nondefective. Clearly, P(A)=7/10, P(B)=7/

1. Sampling with replacement: the object that was drawn at random is placed back to the given set and the set is mixed thoroughly. Then we draw the next object at random. Events A,B are independent. Then we have P(AB)=P(A)P(B)=49% {P(B|A)=P(B)=7/10}

2. Sampling without replacement: the object that was drawn is put aside P(B|A)=6/9. Then we have P(AB)=P(A)P(B|A)=47%

Imagine that a chip is drawn from a box containing 4 chips numbered 000, 011, 101, 110, and let A, B, C be the events that the 1st, 2nd, and 3rd digit, respectively, on the drawn chip is 1

*General case: m events are called independent if P(A 1 ...Am)=P(A 1 )...P(Am) as well as for every k different events Aj1,Aj2,...,Ajk, P(Aj1Aj2...Ajk)=P(Aj1)P(Aj2)...P(Ajk), where k=1,2,...,m-1.

In lecture 16, we learn pairwise independent does not imply Pr(ABC)=Pr(A)Pr(B)Pr(C), here we have another example

Actually, to define the independence of three events, we need: (necessary and sufficient condition) (1) P(AB)=P(A)P(B); (2) P(AC)=P(A)P(C); (3) P(BC)=P(B)P(C); (4) P(ABC)=P(A)P(B)P(C);

In lecture 16, we learn P(AB)=P(A|B)P(B), how about P(ABC)?

P(ABC)=P(AB|C)P(C)

=P(A|BC)P(B|C)P(C)

Also, we can get: P(ABC)=P(CB|A)P(A) =P(C|AB)P(B|A)P(A) etc.

Geometric probability: Romeo and Juliet have a date at a given time, and each will arrive at the meeting place with a delay between 0 and 1 hour, with all pairs of delays being equally likely. The first to arrive will wait for 15 minutes and will leave if the other has not yet arrived. What is the probability that they will meet? (continous-time model)

Let's use x,y to denote the arrival time of Romeo and Juliet, x,y are uniformly choosen from [0,1]. The event they meet (Romeo and Juliet will arrive within 15 minutes of each other) is actually:

x y x y x y

 ^ ^  

Then the shaded area is the event that they meet, the normalized size corresponds to the probability (7/16).