Find the probability distribution of the number of tails in the simultaneous tosses of three coins.
(N/A) When three coins are tossed simultaneously,the sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
Let $X$ represent the number of tails.
It can be seen that $X$ can take the values $0, 1, 2,$ or $3$.
$P(X=0) = P(HHH) = \frac{1}{8}$
$P(X=1) = P(HHT) + P(HTH) + P(THH) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$
$P(X=2) = P(HTT) + P(THT) + P(TTH) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$
$P(X=3) = P(TTT) = \frac{1}{8}$
Thus,the probability distribution is as follows:
Let $X$ represent the number of tails.
It can be seen that $X$ can take the values $0, 1, 2,$ or $3$.
$P(X=0) = P(HHH) = \frac{1}{8}$
$P(X=1) = P(HHT) + P(HTH) + P(THH) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$
$P(X=2) = P(HTT) + P(THT) + P(TTH) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$
$P(X=3) = P(TTT) = \frac{1}{8}$
Thus,the probability distribution is as follows:
| $X$ | $0$ | $1$ | $2$ | $3$ |
| $P(X)$ | $\frac{1}{8}$ | $\frac{3}{8}$ | $\frac{3}{8}$ | $\frac{1}{8}$ |
Explore More
Similar Questions
If $X$ is a Poisson variate such that $\alpha = P(X=1) = P(X=2)$,then $P(X=4)$ is equal to
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 Probability $\omega_{1}$ $\frac{1}{12}$ $\omega_{2}$ $\frac{1}{12}$ $\omega_{3}$ $\frac{1}{6}$ $\omega_{4}$ $\frac{1}{6}$ $\omega_{5}$ $\frac{1}{6}$ $\omega_{6}$ $\frac{3}{2}$
| Outcome | Probability |
| $\omega_{1}$ | $\frac{1}{12}$ |
| $\omega_{2}$ | $\frac{1}{12}$ |
| $\omega_{3}$ | $\frac{1}{6}$ |
| $\omega_{4}$ | $\frac{1}{6}$ |
| $\omega_{5}$ | $\frac{1}{6}$ |
| $\omega_{6}$ | $\frac{3}{2}$ |
Easy
View SolutionIf a variable takes values $0, 1, 2, ..., n$ with frequencies proportional to the binomial coefficients $^nC_0, ^nC_1, ..., ^nC_n$,find the mean of the distribution.
Difficult
View Solution$A$ fair die is tossed twice in succession. If $X$ denotes the number of sixes in two tosses,then the probability distribution of $X$ is given by
The cumulative distribution function $F(X)$ of a discrete random variable $X$ is given by the following table:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $F(X=x)$ $0.2$ $0.37$ $0.48$ $0.62$ $0.85$ $1$
Then $P[X=4] + P[X=5] = $
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
|---|---|---|---|---|---|---|
| $F(X=x)$ | $0.2$ | $0.37$ | $0.48$ | $0.62$ | $0.85$ | $1$ |
Then $P[X=4] + P[X=5] = $
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo