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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

$19$/$6$

Vedclass · Answer

(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 $7.7$.

$X = k$	$-2$	$-1$	$0$	$1$	$2$
$P(X = k)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$

$Z$	$3$	$2$	$1$	$0$	$-1$
$P(Z)$	$0.3$	$0.2$	$0.4$	$0.1$	$0.05$

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(e)$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

$x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x_i)$	$0.2$	$0.5$	$0.15$	$0.25$	$0.1$

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

The variance of the random variable $X$ having the following distribution is:
$X = k$ $-2$ $-1$ $0$ $1$ $2$
$P(X = k)$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$

State which of the following is not the probability distribution of a random variable. Give reasons for your answer.

$Z$ $3$ $2$ $1$ $0$ $-1$

$P(Z)$ $0.3$ $0.2$ $0.4$ $0.1$ $0.05$

Let a sample space be $S = \{\omega_{1}, \omega_{2}, \dots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?
Outcome $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$
$(e)$ $0.1$ $0.2$ $0.3$ $0.4$ $0.5$ $0.6$

$A$ random variable $X$ has the following probability distribution:

$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$

Determine $k$.

If the probability distribution of a random variable $X$ is given by the following table,then $F(0) =$ . . . . . .
$x_i$ $-2$ $-1$ $0$ $1$ $2$
$P(X = x_i)$ $0.2$ $0.5$ $0.15$ $0.25$ $0.1$

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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

The variance of the random variable $X$ having the following distribution is:$X = k$$-2$$-1$$0$$1$$2$$P(X = k)$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{3}$$\frac{1}{6}$$\frac{1}{6}$

State which of the following is not the probability distribution of a random variable. Give reasons for your answer. $Z$ $3$ $2$ $1$ $0$ $-1$ $P(Z)$ $0.3$ $0.2$ $0.4$ $0.1$ $0.05$

Let a sample space be $S = \{\omega_{1}, \omega_{2}, \dots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?Outcome$\omega_1$$\omega_2$$\omega_3$$\omega_4$$\omega_5$$\omega_6$$(e)$$0.1$$0.2$$0.3$$0.4$$0.5$$0.6$

$A$ random variable $X$ has the following probability distribution: $X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$ Determine $k$.

If the probability distribution of a random variable $X$ is given by the following table,then $F(0) =$ . . . . . . $x_i$$-2$$-1$$0$$1$$2$$P(X = x_i)$$0.2$$0.5$$0.15$$0.25$$0.1$