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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