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Suppose that a random variableXhas the uniform distributionon the integers 10, . . . ,20. Find the probability thatXis even.

Short Answer

Expert verified

The probability that X is even is \(\frac{6}{{11}}\) .

Step by step solution

01

Given information

The random variable X follows the uniform distribution on the integers 10,…,20.

02

Determine the probability function

The probability function of any uniformly distributed random variable X is given as,

\(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{{b - a + 1}}\;\;\;for\;x = a,...,b\\0\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)

Since X follows the uniform distribution, the probability function for \(x = 10,...,20\) is,

\(\begin{array}{c}f\left( x \right) = \frac{1}{{20 - 10 + 1}}\\ = \frac{1}{{11}}\end{array}\)

Therefore,

\(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{{11}};\;\;\;\;\;\;\;\;\;\;\;x = 10,...,20\\0\;;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)

03

Obtain the probability

The total even numbers from the integers 10, … , 20 are 6; they are, 10,12,14,16,18, 20.

Since X follows the uniform distribution, this implies that the probability is equally likely for every integer.

The probability that X is even is given as,

\(\begin{array}{c}P\left( {X = 10} \right) + P\left( {X = 12} \right) + P\left( {X = 14} \right) + ... + P\left( {X = 20} \right) = \frac{1}{{11}} + \frac{1}{{11}} + \frac{1}{{11}} + ... + \frac{1}{{11}}\\ = \frac{6}{{11}}\end{array}\)

Therefore, the probability that X is even is \(\frac{6}{{11}}\) .

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