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The discrete random variable π has the shown probability distribution.
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Find the value of π.
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To answer this question, weβll need to recall some information about discrete random variables.
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We know that the sum of all the probabilities for a discrete random variable must be one.
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We can therefore form an equation by adding together the probabilities in our table.
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That gives us π over one plus π over two plus π over three plus π over four is equal to one.
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Next, weβll need to evaluate the expression on the left-hand side of our equation.
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To do that, weβll need to ensure that all of the fractions have the same denominator.
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The lowest common denominator we can choose is found by evaluating the lowest common multiple of one, two, three, and four.
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Itβs 12.
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To change the denominator of our first fraction to 12, weβll need to multiply both the numerator and the denominator by 12.
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That gives us 12π over 12.
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For the second fraction, weβll multiply both the numerator and the denominator by six, to give us six π over 12.
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For π over three, weβll need to multiply both the numerator and the denominator by four, which gives us four π over 12.
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And weβll multiply the numerator and the denominator of π over four by three, to give us three π over 12.
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Next, letβs simplify the expression on the left-hand side.
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Adding together the numerators, 12π plus six π plus four π plus three π, we get 25π.
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So this fraction becomes 25π over 12.
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Then to solve this equation, weβll first multiply both sides by 12, to get 25π equals 12.
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Finally, weβll divide both sides by 25, to get π is equal to 12 25ths.
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Now we can substitute the value of π back into our table showing the probability distribution of the discrete random variable.
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That gives us that the probability π₯ is equal to one is 12 25ths, the probability π₯ is equal to two is six 25ths, the probability is equal to three is four 25ths, and the probability is equal to four is three 25ths.
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Hence, determine the expected value of π.
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Now that we have all the relevant probabilities, we can apply the formula for the expected value of π.
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Itβs the sum of each of the possible outcomes multiplied by the probability of this outcome occurring.
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So letβs substitute what we have into this formula.
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When π₯ is equal to one, it becomes one multiplied by 12 25ths.
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When π₯ is equal to two, π₯ multiplied by the probability of π₯ is two multiplied by six 25ths.
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In the third column, itβs three multiplied by four 25ths.
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And in the fourth and final column, itβs four multiplied by three 25ths.
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That simplifies to 12 25ths plus 12 25ths plus 12 25ths plus another 12 25ths, which is 48 25ths.
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The expected value of π then is 48 25ths.