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
â‡’ 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
â‡’ 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 |
x_{1} |
x_{2} |
x_{3} |
â€¦ |
x_{n} |
P(X) |
p_{1} |
p_{2} |
p_{3} |
â€¦ |
p_{n} |
Where p_{i }> 0, âˆ‘_{i=1}^{n} p_{i} = 1, i = 1,2,3,...,n
Since x_{i} is a possible value of random variable X, and statement "X = x_{i}" is true for some points of the sample space:
P(X = x_{i}) > 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.
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
â‡’ 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
â‡’ 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 |
x_{1} |
x_{2} |
x_{3} |
â€¦ |
x_{n} |
P(X) |
p_{1} |
p_{2} |
p_{3} |
â€¦ |
p_{n} |
Where p_{i }> 0, âˆ‘_{i=1}^{n} p_{i} = 1, i = 1,2,3,...,n
Since x_{i} is a possible value of random variable X, and statement "X = x_{i}" is true for some points of the sample space:
P(X = x_{i}) > 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.