Find the probability distribution of the number of heads in four tosses of a coin.
(N/A) When a coin is tossed four times,the total number of outcomes is $2^4 = 16$. The sample space $S$ is:
$S = \{HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT\}$
Let $X$ be the random variable representing the number of heads. $X$ can take values $0, 1, 2, 3, 4$.
$P(X=0) = P(TTTT) = \frac{1}{16}$
$P(X=1) = P(HTTT) + P(THTT) + P(TTHT) + P(TTTH) = \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$
$P(X=2) = P(HHTT) + P(HTHT) + P(HTTH) + P(THHT) + P(THTH) + P(TTHH) = \frac{6}{16} = \frac{3}{8}$
$P(X=3) = P(HHHT) + P(HHTH) + P(HTHH) + P(THHH) = \frac{4}{16} = \frac{1}{4}$
$P(X=4) = P(HHHH) = \frac{1}{16}$
The probability distribution is:
$S = \{HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT\}$
Let $X$ be the random variable representing the number of heads. $X$ can take values $0, 1, 2, 3, 4$.
$P(X=0) = P(TTTT) = \frac{1}{16}$
$P(X=1) = P(HTTT) + P(THTT) + P(TTHT) + P(TTTH) = \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$
$P(X=2) = P(HHTT) + P(HTHT) + P(HTTH) + P(THHT) + P(THTH) + P(TTHH) = \frac{6}{16} = \frac{3}{8}$
$P(X=3) = P(HHHT) + P(HHTH) + P(HTHH) + P(THHH) = \frac{4}{16} = \frac{1}{4}$
$P(X=4) = P(HHHH) = \frac{1}{16}$
The probability distribution is:
| $X$ | $0$ | $1$ | $2$ | $3$ | $4$ |
| $P(X)$ | $\frac{1}{16}$ | $\frac{1}{4}$ | $\frac{3}{8}$ | $\frac{1}{4}$ | $\frac{1}{16}$ |
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Similar Questions
Let a sample space be $S = \{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?
Outcome $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$ $(a)$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$
| Outcome | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
| $(a)$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ |
Easy
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Number of smart phones $(x)$ $0$ $1$ $2$ $3$ $4$ $5$ Probability $(P(x))$ $k$ $0.3$ $0.15$ $0.15$ $0.1$ $2k$
Then $E(x) = ?$
| Number of smart phones $(x)$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|---|
| Probability $(P(x))$ | $k$ | $0.3$ | $0.15$ | $0.15$ | $0.1$ | $2k$ |
Then $E(x) = ?$
If the probability mass function (p.m.f.) of a random variable $X$ is given by the table below,then find the value of $k$.
$X$ $1$ $2$ $3$ $4$ $5$ $P(X = x)$ $k$ $\frac{k}{3}$ $\frac{k}{4}$ $\frac{k}{2}$ $\frac{k}{2}$
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $P(X = x)$ | $k$ | $\frac{k}{3}$ | $\frac{k}{4}$ | $\frac{k}{2}$ | $\frac{k}{2}$ |
$A$ random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X)$ $0$ $k$ $2k$ $3k$ $3k^2$ $k^2$ $2k^2$ $7k^2+k$
Determine $P(0 < X < 3)$. (in $/10$)
| $X$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |
| $P(X)$ | $0$ | $k$ | $2k$ | $3k$ | $3k^2$ | $k^2$ | $2k^2$ | $7k^2+k$ |
Determine $P(0 < X < 3)$. (in $/10$)
Easy
View SolutionIn a game of throwing $3$ coins, a player loses $₹ 5$ for each head and gains $₹ 10$ for each tail. If a random variable $X: S \rightarrow R$ is defined as $X(a) = \text{net gain } (a \in S)$, then the mean of the random variable is (in rupees):
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