Find the probability distribution of the number of successes in two tosses of a die,where a success is defined as 'six appears on at least one die'.
(N/A) Let $Y$ be the random variable representing the number of successes in two tosses of a die. $A$ success is defined as 'six appears on at least one die'.
The total number of outcomes when two dice are tossed is $6 \times 6 = 36$.
Let $S$ be the event that a six appears on a die,and $F$ be the event that a six does not appear. $P(S) = \frac{1}{6}$ and $P(F) = \frac{5}{6}$.
$P(Y=0)$ is the probability that a six does not appear on either die. This corresponds to the outcomes where neither die shows a six: $P(Y=0) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$.
$P(Y=1)$ is the probability that a six appears on at least one die. This corresponds to the outcomes where at least one die shows a six: $P(Y=1) = 1 - P(Y=0) = 1 - \frac{25}{36} = \frac{11}{36}$.
Wait,let us re-evaluate the definition of success. If success is 'six appears on at least one die',then in two tosses,we can have either $0$ successes or $1$ success (since the event 'six appears on at least one die' is a single outcome event per trial pair). However,if we treat the two tosses as independent trials,the probability distribution is:
The total number of outcomes when two dice are tossed is $6 \times 6 = 36$.
Let $S$ be the event that a six appears on a die,and $F$ be the event that a six does not appear. $P(S) = \frac{1}{6}$ and $P(F) = \frac{5}{6}$.
$P(Y=0)$ is the probability that a six does not appear on either die. This corresponds to the outcomes where neither die shows a six: $P(Y=0) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$.
$P(Y=1)$ is the probability that a six appears on at least one die. This corresponds to the outcomes where at least one die shows a six: $P(Y=1) = 1 - P(Y=0) = 1 - \frac{25}{36} = \frac{11}{36}$.
Wait,let us re-evaluate the definition of success. If success is 'six appears on at least one die',then in two tosses,we can have either $0$ successes or $1$ success (since the event 'six appears on at least one die' is a single outcome event per trial pair). However,if we treat the two tosses as independent trials,the probability distribution is:
| $Y$ | $0$ | $1$ |
| $P(Y)$ | $\frac{25}{36}$ | $\frac{11}{36}$ |
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View Solution$A$ random variable $X$ takes values $0, 1, 2, 3, \ldots$ with probability $P(X=x) = K(x+1)\left(\frac{1}{5}\right)^x$,where $K$ is a constant. Then $P(X=0)$ is:
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 $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$ $(e)$ $0.1$ $0.2$ $0.3$ $0.4$ $0.5$ $0.6$
| Outcome | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
| $(e)$ | $0.1$ | $0.2$ | $0.3$ | $0.4$ | $0.5$ | $0.6$ |
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View Solution$A$ 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
| $X = x_{i}$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
| $P(X = x_{i})$ | $0.1$ | $k$ | $0.2$ | $2k$ | $3k$ | $k$ |
Then the variance of this distribution is
If the probability function of a random variable $X$ is given by $P(X=j) = \frac{1}{2^j}$ for $j = 1, 2, 3, \ldots, \infty$,then the variance of $X$ is:
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