A biased six-sided die is such that $\mathrm{P}\left( X=1 \right)=p$ and $\mathrm{P}\left( X=r+1 \ri

Question

A biased six-sided die is such that $\mathrm{P}\left( X=1 \right)=p$ and $\mathrm{P}\left( X=r+1 \right)=\frac{1}{2}\mathrm{P}\left( X=r \right)$, $r=1$, $2$, $3$, $4$, $5$ where $X$ denotes the number on the uppermost face of the die when it is thrown on the table. Show that $p=\frac{32}{63}$ and find the value of $\mathrm{E}\left( X \right)$. This biased die and an ordinary fair six-sided die are thrown at the same time. If $Y$ denotes the number on the uppermost face of the ordinary die, write down the expectation of $Y$. Find also the value of $\mathrm{P}\left( X+Y=4 \right)$, and $\mathrm{E}\left( X+Y \right)$. Deduce the value of $\mathrm{E}\left( S \right)$, where $S$ denotes the total score of the numbers on the other ten faces of the two dice.

Accepted Answer

$p=\frac{32}{63}$, $E(X)=\frac{40}{21}$, $P(X+Y=4)=\frac4{27}$, $E(X+Y)=\frac{227}{42}$, $E(S)=\frac{1537}{42}$

ACJC Discrete Random Variables Tutorial Q1

ACJC Discrete Random Variables Tutorial Q1

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