Two cards are drawn successively with replace | vedclass

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

Question

(C) Let $p$ be the probability of drawing a ten in a single draw. Since there are $4$ tens in a pack of $52$ cards,$p = \frac{4}{52} = \frac{1}{13}$.T

Vedclass · Accepted Answer

$\frac{2}{13}$

Vedclass · Answer

(C) Let $p$ be the probability of drawing a ten in a single draw. Since there are $4$ tens in a pack of $52$ cards,$p = \frac{4}{52} = \frac{1}{13}$.The probability of not drawing a ten is $q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.Since the cards are drawn with replacement,the number of tens $X$ follows a binomial distribution $B(n, p)$ with $n = 2$ and $p = \frac{1}{13}$.The mean of a binomial distribution is given by $E(X) = np$.Therefore,the mean number of tens is $E(X) = 2 	imes \frac{1}{13} = \frac{2}{13}$.

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. The mean of the number of tens is:

Similar Questions

It is observed that $30 \%$ of the students appearing for a certain entrance test are science students. If $5$ students are randomly selected from this group,the probability of having $2$ science students among these students is

In a Binomial distribution $B(n, p)$,the sum and product of the mean and the variance are $5$ and $6$ respectively,then $6(n+p-q)=$

The probability that a person wins a prize on a lottery ticket is $\frac{1}{4}$. If he purchases $5$ lottery tickets at random,then the probability that he wins at least one prize is

The mean and standard deviation of a binomial variate $X$ are $4$ and $\sqrt{3}$ respectively. Then $P(X \geq 1)$ is equal to

From a lot of $30$ bulbs which include $6$ defectives,a sample of $4$ bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Vedclass Test Series

Exam Paper Generator

Online Exam Module