From a well-shuffled pack of 52 playing cards | vedclass

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

Question

(C) Since the cards are drawn with replacement,each draw is an independent event.Let $E$ be the event of drawing a "king of hearts". The probability o

Vedclass · Accepted Answer

$\frac{51^4}{52^5}$

Vedclass · Answer

(C) Since the cards are drawn with replacement,each draw is an independent event.Let $E$ be the event of drawing a "king of hearts". The probability of drawing a "king of hearts" in any single draw is $P(E) = \frac{1}{52}$.The probability of not drawing a "king of hearts" in any single draw is $P(E') = 1 - \frac{1}{52} = \frac{51}{52}$.We want the $5^{th}$ card to be a "king of hearts". The first $4$ cards can be any card other than the "king of hearts",and the $5^{th}$ card must be the "king of hearts".Probability = $P(E') 	imes P(E') 	imes P(E') 	imes P(E') 	imes P(E) = \left(\frac{51}{52}ight)^4 	imes \frac{1}{52} = \frac{51^4}{52^5}$.

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(e)$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$

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

From a well-shuffled pack of $52$ playing cards,cards are drawn one by one with replacement. The probability that the $5^{th}$ card will be the "king of hearts" is:

Similar Questions

The discrete random variables $X$ and $Y$ are independent from one another and are defined as $X \sim B(16, 0.25)$ and $Y \sim P(2)$. Then the sum of the variances of $X$ and $Y$ is

The random variable $X$ has the following probability distribution:
| $X$ | $8$ | $12$ | $16$ | $20$ | $24$ |
|---|---|---|---|---|---|
| $P(X)$ | $K$ | $\frac{1}{6}$ | $\frac{3}{8}$ | $2K$ | $\frac{1}{12}$ |
Then the value of $K$ is:

Let a sample space be $S = \{\omega_{1}, \omega_{2}, \dots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?
Outcome $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$
$(e)$ $0.1$ $0.2$ $0.3$ $0.4$ $0.5$ $0.6$

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:

In a meeting,$60 \%$ of the members favour and $40 \%$ oppose a certain proposal. $A$ member is selected at random. We define a random variable $X$ such that $X=0$ if the member opposes the proposal and $X=1$ if the member is in favour. Then,$\text{Var}(X) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module