A cubical die with faces marked 1, 2, 3, ..., | vedclass

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(C) Let $E_i$ be the event of getting $i$ on the die. The probability $P(E_i) = k \cdot i^2$ for some constant $k$.Since $\sum_{i=1}^6 P(E_i) = 1$,we

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$\frac{5}{7}$

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(C) Let $E_i$ be the event of getting $i$ on the die. The probability $P(E_i) = k \cdot i^2$ for some constant $k$.Since $\sum_{i=1}^6 P(E_i) = 1$,we have $k(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = 1$.$k(1 + 4 + 9 + 16 + 25 + 36) = 1 \Rightarrow 91k = 1 \Rightarrow k = \frac{1}{91}$.Let $A$ be the event that the number turned up is not even. Then $A = \{1, 3, 5\}$.$P(A) = P(E_1) + P(E_3) + P(E_5) = k(1^2 + 3^2 + 5^2) = k(1 + 9 + 25) = 35k$.We want to find the conditional probability $P(E_5 | A) = \frac{P(E_5 \cap A)}{P(A)}$.Since $E_5 \subset A$,$P(E_5 \cap A) = P(E_5) = 25k$.Thus,$P(E_5 | A) = \frac{25k}{35k} = \frac{25}{35} = \frac{5}{7}$.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$k$	$2k$	$4k$	$2k$	$k$

$A$ cubical die with faces marked $1, 2, 3, ..., 6$ is tossed such that the probability of throwing the number $t$ is proportional to $t^2$. The probability that the number $5$ has appeared,given that the number turned up is not even,is:

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