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Random variables

Random experiment. outcome. numerical measure/aspect of outcome. Random variables. Random variable. Random variable. S. outcome. R. number. Random variables. Random variable is an assignment of a number to an outcome of a random experiment. Notation: Random variable: r.v.

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Random variables

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  1. Random experiment outcome numerical measure/aspect of outcome Random variables Random variable Random variable S outcome R number

  2. Random variables Random variable is an assignment of a number to an outcome of a random experiment. Notation: Random variable: r.v. Example. Select 13 cards from a deck of 52. Random variable X= number of Aces. Possible values of X: 0, 1, 2, 3, 4. X is a discrete r.v. Example. Select a student at random from a class of 200. Random variable X=height of the selected student. Possible values of X: [5ft, 7ft] is a continuous r.v.

  3. DISCRETE RANDOM VARIABLES Discrete r.v. takes on finite (or countable) number of values. What do we need to specify/describe a r.v? Need its values and their probabilities! We describe a discrete r.v. using probability distribution: list of all values of the random variable with the corresponding probabilities. X random variable. Values of X: x1, x2, …., xn Probabilities: P(X= x1 ) =p1, P(X= x2) =p2, …, P(X= xn ) =pn. Every pi≥0 and p1 + p2 + …+ pn =1. Typically, probability distribution of a random variable is given as a table:

  4. Discrete random variables. An example. Toss a fair coin 3 times. Let X be the number of Heads in the 3 tosses. Find the probability distribution of X. Sln. Sample space S={HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. 8 equally likely outcomes. X= # H. Possible values of X: 0, 1, 2, 3. Need probabilities. P(X=0)=P(0 Heads)=P(TTT)=1/8, (X assigned nmbr 1/8 to exprmnt outcome TTT) P(X=1)=P(1 Head)=P(TTH, THT, HTT)=3/8, P(X=2)=P(2 Heads)=P(HHT, HTH, THH)=3/8, P(X=3)=P(3 Heads)=P(HHH)=1/8. Probability distribution of X:

  5. Example, contd. • Find the probability of getting at least 1 Head; • Find the probability of getting 1 or 2 Heads. Sln. Use probability distribution of X. • P(at least 1 Head) = 1 – P( no H) = 1 – P( X=0)=1-1/8=7/8. • P(1 or 2 Heads) = P(X=1 or X=2) = P(X=1) + P(X=2)= 3/8 +3/8=6/8. Complementary events Disjoint events

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