Find the probability distribution of the numb | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 o

Vedclass · Accepted Answer

Vedclass · Answer

(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:

$Y$	$0$	$1$
$P(Y)$	$\frac{25}{36}$	$\frac{11}{36}$

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

$X$	$0$	$1$	$2$
$P(X)$	$\frac{25}{36}$	$k$	$\frac{1}{36}$

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'.

Similar Questions

If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$,whose distribution is given by:
$X=x$ $0$ $1$ $2$ $3$
$P(X=x)$ $\frac{1}{3}$ $\frac{1}{2}$ $0$ $\frac{1}{6}$

Then:

$A$ random variable $X$ has the following probability distribution:
$X$ $1, 2, 3, 4, 5$
$P(X)$ $K^2, 2K, K, 2K, 5K^2$

Then $P(X > 2)$ is equal to:

$A$ coin is tossed three times. Let $X$ denote the number of times a tail follows a head. If $\mu$ and $\sigma^2$ denote the mean and variance of $X$, then the value of $64(\mu+\sigma^2)$ is :

$A$ random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$
$P(X)$ $\frac{25}{36}$ $k$ $\frac{1}{36}$

If the mean of the random variable $X$ is $\frac{1}{3}$,then the variance is:

The probability that an individual suffers a bad reaction from an injection is $0.001$. The probability that out of $2000$ individuals exactly three will suffer a bad reaction is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module