Notes On Probability Distribution of a Random Variable - CBSE Class 12 Maths
Consider six bowls, namely B1, B2, B3, B4, B5 and B6. These bowls have three, five, four, four, six and five marbles, respectively. Let X be the random variable that represents the number of marbles in the bowl selected. X (B1) = 3, X (B2) = 5, X (B3) = 4, X (B4) = 4, X (B5) = 6, X (B6) = 5 Thus, X can take the values, 3, 4, 5 or 6. X = 3, when bowl B1 is selected. X = 4, when bowl B3 or B4 is selected. X = 5, when bowl B2 or B6 is selected. X = 6, when bowl B5 is selected. Assume each bowl is equally likely to be selected. Probability of selecting a bowl = 1/6 P({B1}) = 1/6 ⇒ Probability that X can take value 3 or P(X = 3) = 1/6 P(X = 3) = P({B1}) = 1/6 P({B3, B4}) = 2/6 ⇒ Probability that X can take value 4 or P(X = 4) = 2/6 P(X = 4) = P({B3, B4}) = 2/6 P(X = 5) = P({B2, B6}) = 2/6 P(X = 6) = P({B5}) = 1/6       X      3      4      5      6    P(X)     1/6    2/6    2/6    1/6 The probability distribution of random variable (X) The probability distribution of random variable X is the system of numbers:        X      x1       x2      x3     …     xn     P(X)     p1      p2      p3     …    pn Where pi > 0, ∑i=1n pi = 1, i = 1,2,3,...,n Since xi is a possible value of random variable X, and statement "X = xi" is true for some points of the sample space: P(X = xi) > 0 Random variable X takes all the sample points of a sample space. Hence, the sum of the probabilities in a probability distribution must be one.

#### Summary

Consider six bowls, namely B1, B2, B3, B4, B5 and B6. These bowls have three, five, four, four, six and five marbles, respectively. Let X be the random variable that represents the number of marbles in the bowl selected. X (B1) = 3, X (B2) = 5, X (B3) = 4, X (B4) = 4, X (B5) = 6, X (B6) = 5 Thus, X can take the values, 3, 4, 5 or 6. X = 3, when bowl B1 is selected. X = 4, when bowl B3 or B4 is selected. X = 5, when bowl B2 or B6 is selected. X = 6, when bowl B5 is selected. Assume each bowl is equally likely to be selected. Probability of selecting a bowl = 1/6 P({B1}) = 1/6 ⇒ Probability that X can take value 3 or P(X = 3) = 1/6 P(X = 3) = P({B1}) = 1/6 P({B3, B4}) = 2/6 ⇒ Probability that X can take value 4 or P(X = 4) = 2/6 P(X = 4) = P({B3, B4}) = 2/6 P(X = 5) = P({B2, B6}) = 2/6 P(X = 6) = P({B5}) = 1/6       X      3      4      5      6    P(X)     1/6    2/6    2/6    1/6 The probability distribution of random variable (X) The probability distribution of random variable X is the system of numbers:        X      x1       x2      x3     …     xn     P(X)     p1      p2      p3     …    pn Where pi > 0, ∑i=1n pi = 1, i = 1,2,3,...,n Since xi is a possible value of random variable X, and statement "X = xi" is true for some points of the sample space: P(X = xi) > 0 Random variable X takes all the sample points of a sample space. Hence, the sum of the probabilities in a probability distribution must be one.

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