91Ó°ÊÓ

The value of$P\left\{X_6>X_1âˆ\pounds X_1=\max \left(X_1,…,X_5\right)\right\}$

Let $X_1,X_2.....$be continuous random variables with independent and identical distributions.

Conditional probability is defined as follows:

$\begin{array}{r}P\left\{X_6>X_1âˆ\pounds X_1=max\left(X_1,…,X_5\right)\right\}=\frac{P\left\{X_6>X_1,X_1=max\left(X_1,…,X_5\right)\right\}}{P\left\{X_1=max\left(X_1,…,X_5\right)\right\}}\\ =\frac{P\left\{\left(X_6>X_1\right)âˆ\copyright \left[X_1=max\left(X_1,…,X_5\right)\right]\right\}}{P\left\{X_1=max\left(X_1,…,X_5\right)\right\}}\\ ≈\frac{P\left\{\left(X_6=max\left(X_1,…,X_6\right)\right)âˆ\copyright \left[X_1=max\left(X_1,…,X_5\right)\right]\right\}}{P\left\{X_1=max\left(X_1,…,X_5\right)\right\}}\\ ≈\frac{P\left(X_6=max\left(X_1,…,X_6\right)\right)×P\left(X_1=max\left(X_1,…,X_5\right)\right)}{P\left\{X_1=max\left(X_1,…,X_5\right)\right\}}\\ =\frac{\left(\frac{1}{6}\right)\left(\frac{1}{5}\right)}{\left(\frac{1}{5}\right)}\\ =5â‹\dots \left(\frac{1}{6}×\frac{1}{5}\right)\\ =\frac{1}{6}\end{array}$

Hence$P\left\{X_6>X_1âˆ\pounds X_1=\max \left(X_1,…,X_5\right)\right\}\text{is}\frac{1}{6}$

The value of$P\left\{X_6>X_2âˆ\pounds X_1=\max \left(X_1,…,X_5\right)\right\}$

Given: $P\left\{X_6>X_2âˆ\pounds X_1=\max \left(X_1,…,X_5\right),X_6>X_1\right\}$

Calculation: Let $X_1,X_2.....$be continuous random variables with independent and identical distributions.

Where $X_6>X_1$

$P\left\{X_6>X_2âˆ\pounds X_1=\max \left(X_1,…,X_5\right),X_6>X_1\right\}=1$

By symmetry,

$P\left\{X_6>X_2âˆ\pounds X_1=\max \left(X_1,…,X_5\right),X_6<X_1\right\}=\frac{1}{2}$

From part (a),

$P\left\{X_6>X_1âˆ\pounds X_1=\max \left(X_1,…,X_5\right)\right\}=\frac{1}{6}$

Therefore

$\begin{array}{r}P\left\{X_6>X_2âˆ\pounds X_1=max\left(X_1,…,X_5\right)\right\}=\frac{1}{6}+\left(\frac{1}{2}×\frac{5}{6}\right)\\ =\frac{7}{12}\end{array}$

Hence$P\left\{X_6>X_2âˆ\pounds X_1=\max \left(X_1,…,X_5\right)\right\}=\frac{7}{12}$