The king,queen,and jack of clubs are removed from a deck of $52$ playing cards and then well shuffled. Now,one card is drawn at random from the remaining cards. Determine the probability that the card is:
$(i)$ a heart
$(ii)$ a king
$(i)$ a heart
$(ii)$ a king
(A) Total number of cards in a deck = $52$.
After removing the king,queen,and jack of clubs,the number of remaining cards is $n(S) = 52 - 3 = 49$.
$(i)$ Let $E_1$ be the event of getting a heart.
Since there are $13$ hearts in a deck and none were removed,$n(E_1) = 13$.
Therefore,the probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{13}{49}$.
$(ii)$ Let $E_2$ be the event of getting a king.
There are $4$ kings in a deck. Since the king of clubs was removed,the number of remaining kings is $n(E_2) = 4 - 1 = 3$.
Therefore,the probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{3}{49}$.
After removing the king,queen,and jack of clubs,the number of remaining cards is $n(S) = 52 - 3 = 49$.
$(i)$ Let $E_1$ be the event of getting a heart.
Since there are $13$ hearts in a deck and none were removed,$n(E_1) = 13$.
Therefore,the probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{13}{49}$.
$(ii)$ Let $E_2$ be the event of getting a king.
There are $4$ kings in a deck. Since the king of clubs was removed,the number of remaining kings is $n(E_2) = 4 - 1 = 3$.
Therefore,the probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{3}{49}$.
Explore More
Similar Questions
$A$ circle with radius $7 \, cm$ is drawn at the centre of a rectangle $28 \, cm$ long and $22 \, cm$ broad. Find the probability that a ball hitting the rectangle hits the circle.
Difficult
View Solution$A$ box contains $5$ red marbles,$8$ white marbles and $4$ green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
$(i)$ red?
$(ii)$ white?
$(iii)$ not green?
$(i)$ red?
$(ii)$ white?
$(iii)$ not green?
Difficult
View Solution$A$ box contains $12$ balls out of which $x$ are black. If one ball is drawn at random from the box,what is the probability that it will be a black ball?
If $6$ more black balls are put in the box,the probability of drawing a black ball is now double of what it was before. Find $x$.
If $6$ more black balls are put in the box,the probability of drawing a black ball is now double of what it was before. Find $x$.
Medium
View SolutionTwo unbiased dice are thrown once. Write all the possible outcomes of this experiment. What is the probability that the sum of numbers on two dice is a factor of $18$?
Medium
View SolutionWhich of the following arguments are correct and which are not correct? Give reasons for your answer.
$(i)$ If two coins are tossed simultaneously,there are three possible outcomes - two heads,two tails,or one of each. Therefore,for each of these outcomes,the probability is $\frac{1}{3}$.
$(ii)$ If a die is thrown,there are two possible outcomes - an odd number or an even number. Therefore,the probability of getting an odd number is $\frac{1}{2}$.
$(i)$ If two coins are tossed simultaneously,there are three possible outcomes - two heads,two tails,or one of each. Therefore,for each of these outcomes,the probability is $\frac{1}{3}$.
$(ii)$ If a die is thrown,there are two possible outcomes - an odd number or an even number. Therefore,the probability of getting an odd number is $\frac{1}{2}$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo