(A) The total number of cards is $52$. The number of spade cards is $13$.Probability of drawing a spade in one draw,$P(S) = \frac{13}{52} = \frac{1}{4

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

(A) The total number of cards is $52$. The number of spade cards is $13$.Probability of drawing a spade in one draw,$P(S) = \frac{13}{52} = \frac{1}{4}$.Probability of not drawing a spade in one draw,$P(S') = 1 - \frac{1}{4} = \frac{3}{4}$.Since the card is replaced each time,the events are independent.The probability that he fails the first two times means he does not get a spade in the first draw $AND$ he does not get a spade in the second draw.Required probability $= P(S') \times P(S') = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$.

Class 12 Mathematics Probability Probability distribution

Medium

$A$ man draws a card from a pack of $52$ playing cards,replaces it,and shuffles the pack. He continues this process until he gets a spade card. The probability that he will fail the first two times is:

A
$\frac{9}{16}$
B
$\frac{1}{16}$
C
$\frac{9}{64}$
D
None of these

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Probability

Subtopic

Probability distribution

If the mean of the following probability distribution of a random variable $X$ is $\frac{46}{9}$,then the variance of the distribution is:

$X$ $0$ $2$ $4$ $6$ $8$

$P(X)$ $a$ $2a$ $a+b$ $2b$ $3b$

DifficultJEE MAIN 2024

View Solution

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:

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

MediumTS EAMCET 2009

View Solution

When an unfair dice is thrown,the probability of getting a number $k$ on it is $P(X=k)=k^2 P$,where $k=1, 2, 3, 4, 5, 6$ and $X$ is the random variable denoting a number on the dice,then the mean of $X$ is

EasyAP EAMCET 2024

View Solution

In a game,$3$ coins are tossed. $A$ person is paid ₹ $100$,if he gets all heads or all tails; and he is supposed to pay ₹ $40$,if he gets one head or two heads. The amount he can expect to win/lose on an average per game in (₹) is

MediumMHT CET 2024

View Solution

$A$ random variable $X$ has the following probability distribution:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X = x)$ $0.15$ $0.23$ $k$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

For the events $E = \{x : x \text{ is a prime number}\}$ and $F = \{x : x < 4\}$, then $P(E \cup F) = $

MediumTS EAMCET 2020

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$k$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

$A$ man draws a card from a pack of $52$ playing cards,replaces it,and shuffles the pack. He continues this process until he gets a spade card. The probability that he will fail the first two times is:

Similar Questions

If the mean of the following probability distribution of a random variable $X$ is $\frac{46}{9}$,then the variance of the distribution is: $X$ $0$ $2$ $4$ $6$ $8$ $P(X)$ $a$ $2a$ $a+b$ $2b$ $3b$

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:

When an unfair dice is thrown,the probability of getting a number $k$ on it is $P(X=k)=k^2 P$,where $k=1, 2, 3, 4, 5, 6$ and $X$ is the random variable denoting a number on the dice,then the mean of $X$ is

In a game,$3$ coins are tossed. $A$ person is paid ₹ $100$,if he gets all heads or all tails; and he is supposed to pay ₹ $40$,if he gets one head or two heads. The amount he can expect to win/lose on an average per game in (₹) is

$A$ random variable $X$ has the following probability distribution:$X = x$$1$$2$$3$$4$$5$$6$$7$$8$$P(X = x)$$0.15$$0.23$$k$$0.10$$0.20$$0.08$$0.07$$0.05$For the events $E = \{x : x \text{ is a prime number}\}$ and $F = \{x : x < 4\}$, then $P(E \cup F) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If the mean of the following probability distribution of a random variable $X$ is $\frac{46}{9}$,then the variance of the distribution is:

$X$ $0$ $2$ $4$ $6$ $8$

$P(X)$ $a$ $2a$ $a+b$ $2b$ $3b$

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:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X = x)$ $0.15$ $0.23$ $k$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

For the events $E = \{x : x \text{ is a prime number}\}$ and $F = \{x : x < 4\}$, then $P(E \cup F) = $