From a pack of 52 cards, two cards are drawn one by one without replacement. The probability that first drawn card is a king and second is a queen, is
A.\[ \dfrac{2}{13}\]
B. \[ \dfrac{8}{663}\]
C. \[ \dfrac{4}{663}\]
D. \[ \dfrac{103}{663}\]
Answer
266.4k+ views
Hint: We all have a basic idea of cards. There are 52 cards in a pack. Among them, there are 4 types of cards such as 13 black spades, 13 red hearts, 13 blue diamonds and 13 green clubs.
Formula Used: Let E be an event then probability of E = P(E)= Number of favourable outcomes of E/Total number of outcomes of E.
Complete step by step solution:We all have a basic idea of cards. There are 52 cards in a pack. Among them, there are 4 types of cards such as 13 black spades, 13 red hearts, 13 blue diamonds and 13 green clubs. There are 4 king cards and 4 queen cards.
If we draw one king card among 52 cards one king card can be drawn in \(\dfrac{4}{{52}}\) ways and for each of drawn, one queen card can be drawn in \(\dfrac{4}{{51}}\)as one card has been drawn already (52-1=51, but there are 4 queen cards) .
Therefore, the probability that first drawn card is a king and second is a queen = \(\dfrac{4}{{52}}.\dfrac{4}{{51}} = \dfrac{4}{{663}}\) .
Option ‘C’ is correct
Note: This type of problem is very easy if you have the knowledge of the definition of probability and basic idea of pack of cards.
Formula Used: Let E be an event then probability of E = P(E)= Number of favourable outcomes of E/Total number of outcomes of E.
Complete step by step solution:We all have a basic idea of cards. There are 52 cards in a pack. Among them, there are 4 types of cards such as 13 black spades, 13 red hearts, 13 blue diamonds and 13 green clubs. There are 4 king cards and 4 queen cards.
If we draw one king card among 52 cards one king card can be drawn in \(\dfrac{4}{{52}}\) ways and for each of drawn, one queen card can be drawn in \(\dfrac{4}{{51}}\)as one card has been drawn already (52-1=51, but there are 4 queen cards) .
Therefore, the probability that first drawn card is a king and second is a queen = \(\dfrac{4}{{52}}.\dfrac{4}{{51}} = \dfrac{4}{{663}}\) .
Option ‘C’ is correct
Note: This type of problem is very easy if you have the knowledge of the definition of probability and basic idea of pack of cards.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Mutually Exclusive vs Independent Events: Key Differences Explained

Area vs Volume: Key Differences Explained for Students

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

Sign up for JEE Main 2026 Live Classes - Vedantu

JEE Main 2026 Helpline Numbers - Center Contact, Phone Number, Address

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Marks vs Rank: Estimate IIT Rank from Your Score

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Electromagnetic Waves and Their Importance

