$A$ bag contains $2$ white and $1$ red balls. One ball is drawn at random and then put back in the box after noting its colour. The process is repeated again. If $X$ denotes the number of red balls recorded in the two draws,describe $X$.
(N/A) Let the balls in the bag be denoted by $w_{1}, w_{2}, r$. The sample space $S$ for two draws with replacement is given by:
$S = \{w_{1}w_{1}, w_{1}w_{2}, w_{2}w_{1}, w_{2}w_{2}, w_{1}r, w_{2}r, rw_{1}, rw_{2}, rr\}$
Here,$X$ is a random variable representing the number of red balls in the two draws.
For outcomes where no red ball is drawn: $X(w_{1}w_{1}) = X(w_{1}w_{2}) = X(w_{2}w_{1}) = X(w_{2}w_{2}) = 0$.
For outcomes where exactly one red ball is drawn: $X(w_{1}r) = X(w_{2}r) = X(rw_{1}) = X(rw_{2}) = 1$.
For the outcome where two red balls are drawn: $X(rr) = 2$.
Thus,$X$ is a random variable that can take values $0, 1,$ or $2$.
$S = \{w_{1}w_{1}, w_{1}w_{2}, w_{2}w_{1}, w_{2}w_{2}, w_{1}r, w_{2}r, rw_{1}, rw_{2}, rr\}$
Here,$X$ is a random variable representing the number of red balls in the two draws.
For outcomes where no red ball is drawn: $X(w_{1}w_{1}) = X(w_{1}w_{2}) = X(w_{2}w_{1}) = X(w_{2}w_{2}) = 0$.
For outcomes where exactly one red ball is drawn: $X(w_{1}r) = X(w_{2}r) = X(rw_{1}) = X(rw_{2}) = 1$.
For the outcome where two red balls are drawn: $X(rr) = 2$.
Thus,$X$ is a random variable that can take values $0, 1,$ or $2$.
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View SolutionThe following table represents the probability distribution of a random variable $X$ for some $k \in Q$. Find the value of $k$.
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| $X=x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
| $P(X=x)$ | $0.1$ | $k$ | $0.2$ | $2k$ | $0.3$ | $k$ |
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The cumulative distribution function (c.d.f.) $F(x)$ of a discrete random variable $X$ is given by the following table:
$X$ $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$ $F(X)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Then,find $P[X=3]$.
| $X$ | $-3$ | $-1$ | $0$ | $1$ | $3$ | $5$ | $7$ | $9$ |
|---|---|---|---|---|---|---|---|---|
| $F(X)$ | $0.1$ | $0.3$ | $0.5$ | $0.65$ | $0.75$ | $0.85$ | $0.90$ | $1$ |
Then,find $P[X=3]$.
If the probability function of a random variable $X$ is given by $P(X=k) = \frac{3^{ck}}{k!}$ for $k = 1, 2, 3, \ldots$ (where $c$ is a constant),then $c =$
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