Consider a game of tossing a six-sided fair d | vedclass

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(A) Let $X$ be the random variable representing the winning amount. The probability of each face $k$ appearing is $P(k) = 1/6$ for $k \in \{1, 2, 3, 4

Vedclass · Accepted Answer

$19$/$6$

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(A) Let $X$ be the random variable representing the winning amount. The probability of each face $k$ appearing is $P(k) = 1/6$ for $k \in \{1, 2, 3, 4, 5, 6\}$.The winning amount for a face $k$ is $W(k) = \frac{36}{k} - 2k$.The expected value $E[X]$ is given by $\sum_{k=1}^{6} W(k) \cdot P(k)$.$E[X] = \frac{1}{6} \sum_{k=1}^{6} (\frac{36}{k} - 2k) = \frac{1}{6} [(\frac{36}{1} - 2) + (\frac{36}{2} - 4) + (\frac{36}{3} - 6) + (\frac{36}{4} - 8) + (\frac{36}{5} - 10) + (\frac{36}{6} - 12)]$.$E[X] = \frac{1}{6} [34 + 14 + 6 + 1 + (7.2 - 10) + (6 - 12)]$.$E[X] = \frac{1}{6} [55 - 2.8 - 6] = \frac{1}{6} [46.2] = 7.7$.Since the provided options do not match the calculated value, the question is flawed. Assuming the sum was intended to be over $k \in \{1, 2, 3, 4, 5, 6\}$ and the expression was $\sum (\frac{36}{k} - 2k)$, the result is $7.7$.

$X = x_{i}$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x_{i})$	$0.1$	$k$	$0.2$	$2k$	$3k$	$k$

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

Consider a game of tossing a six-sided fair die. If the face that comes up is $k$, the player wins Rs. $36/k$ and loses Rs. $2k$, where $k \in \{1, 2, 3, 4, 5, 6\}$. What is the expected winning amount in this game in Rs.?

Similar Questions

$A$ random variable $X$ has the following distribution.
$X = x_{i}$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X = x_{i})$ $0.1$ $k$ $0.2$ $2k$ $3k$ $k$

Then the variance of this distribution is

$A$ fair $n$-faced die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9}$,then $n=$ (where $n \in N$ ).

For the following probability distribution,the standard deviation of the random variable $X$ is:

$X$ $2$ $3$ $4$

$P(X=x)$ $0.2$ $0.5$ $0.3$

If the following function is a probability density function of a random variable $X$, $f(x) = kx^2(1 - x)$ for $0 < x < 1$ and $f(x) = 0$ otherwise, then the value of $k$ is:

The probability distribution of a discrete random variable $X$ is given by the following table:
$X = x$ $0$ $1$ $2$ $3$ $4$
$P(X = x)$ $k$ $2k$ $4k$ $2k$ $k$

Then the value of $P(X \leq 2)$ is:

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