Two bad eggs are mixed accidentally with 10 g | vedclass

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

Question

(B) Let $X$ be the random variable representing the number of bad eggs drawn. The total number of eggs is $12$, and we draw $3$ eggs without replaceme

Vedclass · Accepted Answer

$\frac{15}{44}$

Vedclass · Answer

(B) Let $X$ be the random variable representing the number of bad eggs drawn. The total number of eggs is $12$, and we draw $3$ eggs without replacement. The possible values for $X$ are $0, 1, 2$.The probability distribution is calculated as follows:$P(X=0) = \frac{{}^{10}C_3}{{}^{12}C_3} = \frac{120}{220} = \frac{12}{22}$$P(X=1) = \frac{{}^{10}C_2 	imes {}^{2}C_1}{{}^{12}C_3} = \frac{45 	imes 2}{220} = \frac{90}{220} = \frac{9}{22}$$P(X=2) = \frac{{}^{10}C_1 	imes {}^{2}C_2}{{}^{12}C_3} = \frac{10 	imes 1}{220} = \frac{10}{220} = \frac{1}{22}$Now, we calculate the mean $E(X) = \sum x_i P_i$:$E(X) = (0 	imes \frac{12}{22}) + (1 	imes \frac{9}{22}) + (2 	imes \frac{1}{22}) = \frac{9+2}{22} = \frac{11}{22} = \frac{1}{2}$Next, we calculate $E(X^2) = \sum x_i^2 P_i$:$E(X^2) = (0^2 	imes \frac{12}{22}) + (1^2 	imes \frac{9}{22}) + (2^2 	imes \frac{1}{22}) = \frac{9+4}{22} = \frac{13}{22}$Finally, the variance is given by $Var(X) = E(X^2) - [E(X)]^2$:$Var(X) = \frac{13}{22} - (\frac{1}{2})^2 = \frac{13}{22} - \frac{1}{4} = \frac{26-11}{44} = \frac{15}{44}$

$X = x$	$0$	$1$	$2$
$P(X = x)$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{2}{5}$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{2}{20}$	$\frac{4}{20}$	$\frac{6}{20}$	$\frac{8}{20}$

Two bad eggs are mixed accidentally with $10$ good ones. If three eggs are drawn at random from this lot in succession without replacement, then the variance of the probability distribution of the number of bad eggs drawn is

Similar Questions

If a random variable $X$ denotes the number that appears on the upper face of a die when it is rolled,then $\frac{\text{Variance of } X}{\text{Mean of } X}$ is equal to

The probability distribution of a random variable $X$ is given by
$X = x$ $0$ $1$ $2$
$P(X = x)$ $\frac{1}{5}$ $\frac{2}{5}$ $\frac{2}{5}$

Then the variance of $X$ is

If the probability distribution of a random variable $X$ is as follows, then the variance of $X$ is
$\begin{array}{|c|c|c|c|c|}\hline X=x & 2 & 3 & 5 & 9 \\\hline P(X=x) & K & 2 K & 3 K^2 & K^2 \\\hline\end{array}$

The distribution of a random variable $X$ is given below:
$X=x$ $1$ $2$ $3$ $4$
$P(X=x)$ $\frac{2}{20}$ $\frac{4}{20}$ $\frac{6}{20}$ $\frac{8}{20}$

Then,the standard deviation of $X$ is

If the probability of a bad reaction from a vaccination is $0.01$, then the probability that exactly two out of $300$ people will get a bad reaction is

Vedclass Test Series

Exam Paper Generator

Online Exam Module