Which of the following can not be a valid assignment of probabilities for outcomes of sample space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$?
Outcome Probability $\omega_{1}$ $0.1$ $\omega_{2}$ $0.01$ $\omega_{3}$ $0.05$ $\omega_{4}$ $0.03$ $\omega_{5}$ $0.01$ $\omega_{6}$ $0.2$ $\omega_{7}$ $0.6$
| Outcome | Probability |
| $\omega_{1}$ | $0.1$ |
| $\omega_{2}$ | $0.01$ |
| $\omega_{3}$ | $0.05$ |
| $\omega_{4}$ | $0.03$ |
| $\omega_{5}$ | $0.01$ |
| $\omega_{6}$ | $0.2$ |
| $\omega_{7}$ | $0.6$ |
(D) For an assignment of probabilities to be valid,it must satisfy two conditions:
$1$. Each probability $p(\omega_{i})$ must be such that $0 \leq p(\omega_{i}) \leq 1$.
$2$. The sum of all probabilities must be equal to $1$,i.e.,$\sum_{i=1}^{7} p(\omega_{i}) = 1$.
In the given case:
Sum $= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.0$.
Since the sum is $1$ and each individual probability is between $0$ and $1$,this is a valid assignment of probabilities.
$1$. Each probability $p(\omega_{i})$ must be such that $0 \leq p(\omega_{i}) \leq 1$.
$2$. The sum of all probabilities must be equal to $1$,i.e.,$\sum_{i=1}^{7} p(\omega_{i}) = 1$.
In the given case:
Sum $= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.0$.
Since the sum is $1$ and each individual probability is between $0$ and $1$,this is a valid assignment of probabilities.
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$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X = x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$
Then, $P(0 < X < 4)$ is equal to:
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| $P(X = x)$ | $0$ | $k$ | $2k$ | $2k$ | $3k$ | $k^2$ | $2k^2$ | $7k^2 + k$ |
Then, $P(0 < X < 4)$ is equal to:
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