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| Random Variable |
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A variable with numeric values that can be determined from a random phenomenon.
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- For example, the random variable is the number on the top surface of a dice, namely \(\lbrace1, 2, 3, 4, 5, 6 \rbrace\)
- There are two types of random variables, namely discrete random variables and continuous random variables
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| Discrete |
Continuous |
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Random variables that have countable numbers of values, usually taking values like zero and positive integers
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Random variables that are not integers but take values that lie in an interval
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\(X = \lbrace r : r = 0, 1, 2, 3\rbrace\) |
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\(Y = \lbrace y : y \text{ is the pupil's height in cm}, \ a \leqslant y \leqslant b \rbrace\) |
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- The probability of each event can be summarised in the probability distribution table for \(X\):
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If \(X\) is a discrete random variable with the values
\(r_1, r_2, r_3, ..., r_n \) and their respective probabilities are
\(P(X = r_1), P(X = r_2), P(X = r_3), ..., P(X = r_n),\) then,
\(\sum_{i=1}^{n}\ P(X = r_i) = 1 \), thus each \(P(X = r_i) \geqslant 0.\)
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- Other than tree diagram, the probability distribution for each discrete random variable \(X\) can be represented by a table and a graph
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The choice of the second or the third product is not dependent on the choice of the first product as the earlier product has been returned to the box. These are independent events.
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| Example |
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\(70\%\) of Form \(5\) Dahlia pupils achieved a grade A in the final year examination for the science subject.
Two pupils were chosen at random from that class.
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If \(X\) represents the number of pupils who did not get a grade A, construct a table to show all the possible values of \(X\) with their corresponding probabilities.
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Next, draw a graph to show the probability distribution of \(X\).
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Solution:
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\(P (A : A \text{ is a pupil who did not achieve a grade A}) \\ = 1-\dfrac{70}{100}\\ = 0.3\) |
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\(P (B : B \text{ is a pupil who achieved a grade A}) \\ = \dfrac{70}{100}\\ = 0.7\) |
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Then, \(X = \{0, 1, 2\}\) |
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\(\begin{aligned} P(X = 0) &= P(B, B) \\ &= 0.7 × 0.7\\ &=0.49\\\\ P(X = 1) &= P(A, B) + P(B, A)\\ &= (0.3 × 0.7) + (0.7 × 0.3)\\ &=0.42\\\\ P(X = 2) &= P(A, A) \\ &= 0.3 × 0.3\\ &=0.09 \end{aligned}\) |
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| \(X=r\) |
\(0\) |
\(1\) |
\(2\) |
| \(P(X=r)\) |
\(0.49\) |
\(0.42\) |
\(0.09\) |
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| (b) |
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