Five cards are drawn successively with replac | vedclass

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

Question

(A) Let $X$ represent the number of spade cards among the five cards drawn. Since the drawing of cards is with replacement,the trials are Bernoulli tr

Vedclass · Accepted Answer

$\frac{1}{1024}$

Vedclass · Answer

(A) Let $X$ represent the number of spade cards among the five cards drawn. Since the drawing of cards is with replacement,the trials are Bernoulli trials.In a well-shuffled deck of $52$ cards,there are $13$ spade cards.Therefore,the probability of drawing a spade in a single trial is $p = \frac{13}{52} = \frac{1}{4}$.The probability of not drawing a spade is $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$.$X$ follows a binomial distribution with $n = 5$ and $p = \frac{1}{4}$.The probability mass function is given by $P(X = x) = ^{n}C_{x} q^{n-x} p^{x}$,where $x = 0, 1, 2, 3, 4, 5$.We want to find the probability that all five cards are spades,which is $P(X = 5)$.$P(X = 5) = ^{5}C_{5} \left(\frac{3}{4}\right)^{5-5} \left(\frac{1}{4}\right)^{5}$.$P(X = 5) = 1 \cdot \left(\frac{3}{4}\right)^{0} \cdot \left(\frac{1}{4}\right)^{5}$.$P(X = 5) = 1 \cdot 1 \cdot \frac{1}{1024} = \frac{1}{1024}$.

Five cards are drawn successively with replacement from a well-shuffled deck of $52$ cards. What is the probability that all the five cards are spades?

Similar Questions

$A$ lot of $100$ bulbs contains $10$ defective bulbs. Five bulbs are selected at random from the lot and sent to a retail store. The probability that the store will receive at most one defective bulb is:

One out of $9$ ships is likely to sink when they are set on sail. When $6$ ships are set on sail,the probability that exactly $3$ of them will not arrive safely is

For $X \sim B(n, p)$,if $p=0.6$ and $E(X)=6$,then $\operatorname{Var}(X)=$

The mean and variance of a binomial distribution are $4$ and $3$ respectively. Then,the probability of getting exactly six successes in this distribution is:

The random variable $X$ follows a binomial distribution $B(n, p)$ for which the difference between the mean and the variance is $1$. If $2 P(X=2) = 3 P(X=1)$,then $n^2 P(X > 1)$ is equal to $......$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module