$Let X_{1}, X_{2} and X_{3}$ be independent and identically distributed random variables with the uniform distribution on [0, 1]. The probability $p {x_{1} is the largest} is$

0.33

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Solution

The correct option is A 0.33
$As X_{1}, X_{2} and X_{3}$ are independent and identically distributed so $P (x_{1}) = P (x_{2}) = P (x_{3})$

$So P (x_{1} is largest) = P (X_{2} is largest) = P (x_{3} is largest)$

$= \frac{1}{3} = 0.33$

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Let X1, X2 and X3 be independent and identically distributed random variables with the uniform distribution on [0, 1]. The probability p{x1 is the largest} is 0.33

The correct option is A 0.33As X1, X2 and X3 are independent and identically distributed so P(x1)=P(x2)=P(x3) So P(x1 is largest)=P(X2 is largest)=P(x3 is largest) =13=0.33

$Let X_{1}, X_{2} and X_{3}$ be independent and identically distributed random variables with the uniform distribution on [0, 1]. The probability $p {x_{1} is the largest} is$

0.33

The correct option is A 0.33
$As X_{1}, X_{2} and X_{3}$ are independent and identically distributed so $P (x_{1}) = P (x_{2}) = P (x_{3})$

$So P (x_{1} is largest) = P (X_{2} is largest) = P (x_{3} is largest)$

$= \frac{1}{3} = 0.33$