$A$ die is thrown repeatedly until a six comes up. What is the sample space for this experiment?
In this experiment,the die is thrown until a $6$ appears. Let $S$ denote the sample space.
If $6$ appears on the first throw,the outcome is $6$.
If $6$ appears on the second throw,the outcome is $(x, 6)$ where $x \in \{1, 2, 3, 4, 5\}$.
If $6$ appears on the third throw,the outcome is $(x, y, 6)$ where $x, y \in \{1, 2, 3, 4, 5\}$.
Continuing this process,the sample space $S$ is given by:
$S = \{6, (x, 6), (x, y, 6), (x, y, z, 6), \dots \}$ where $x, y, z, \dots \in \{1, 2, 3, 4, 5\}$.
If $6$ appears on the first throw,the outcome is $6$.
If $6$ appears on the second throw,the outcome is $(x, 6)$ where $x \in \{1, 2, 3, 4, 5\}$.
If $6$ appears on the third throw,the outcome is $(x, y, 6)$ where $x, y \in \{1, 2, 3, 4, 5\}$.
Continuing this process,the sample space $S$ is given by:
$S = \{6, (x, 6), (x, y, 6), (x, y, z, 6), \dots \}$ where $x, y, z, \dots \in \{1, 2, 3, 4, 5\}$.
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$A$ random variable $X$ has the following probability distribution:
| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(x)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{x \text{ is a prime number}\}$ and $F = \{x < 4\}$,the probability $P(E \cup F)$ is:
| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(x)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{x \text{ is a prime number}\}$ and $F = \{x < 4\}$,the probability $P(E \cup F)$ is:
DifficultMHT CET 2008
View Solution$A$ random variable $X$ takes values $0, 1, 2, 3, \dots$ with probabilities $P(X=x) = k(x+1)\left(\frac{1}{2}\right)^x$. If $k$ is a constant,then $P(X=1) = $
If the probability mass function (p.m.f.) of a random variable $X$ is given by the table below,then find the value of $k$.
$X$ $1$ $2$ $3$ $4$ $5$ $P(X = x)$ $k$ $\frac{k}{3}$ $\frac{k}{4}$ $\frac{k}{2}$ $\frac{k}{2}$
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $P(X = x)$ | $k$ | $\frac{k}{3}$ | $\frac{k}{4}$ | $\frac{k}{2}$ | $\frac{k}{2}$ |
$A$ die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die,then $P(E)$ is equal to:
If $X$ is a Poisson variate such that $\alpha = P(X=1) = P(X=2)$, then $P(X=4)$ is equal to
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