49 votes 49 votes A deck of $5$ cards (each carrying a distinct number from $1$ to $5$) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card? $\left(\dfrac{1}{5}\right)$ $\left(\dfrac{4}{25}\right)$ $\left(\dfrac{1}{4}\right)$ $\left(\dfrac{2}{5}\right)$ Probability gatecse-2011 probability normal + – go_editor asked Sep 29, 2014 • edited Nov 30, 2017 by pavan singh go_editor 18.2k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments sanjaysharmarose commented Jun 25, 2020 i edited by sanjaysharmarose Jun 25, 2020 reply Follow Share 5C2 is 10 . So your answer will be 4/10 = 2/5 which is wrong. 5C2 means pick two items out of 5 and here (1,2) will be same as (2,1) . You can use 5P2 or 5C1*4C1 = 20 . 7 votes 7 votes Paatni22 commented Dec 28, 2020 reply Follow Share yes bcoz we have to take the ordered pairs in SS instead of unordered pairs here, order matters here. 0 votes 0 votes Ray Tomlinson commented Sep 30, 2023 i edited by Arjun Sep 30, 2023 reply Follow Share why total possiblity is , 20 not 25 ?If anyone is confused about how sample space is 20 and not 25. The reason is question is considering “without replacement”. Hence there are 5 ways to choose 1st card and 4 ways to choose the 2nd card ,hence the |SS|=5x4=20 0 votes 0 votes Please log in or register to add a comment.
Best answer 65 votes 65 votes The number on the first card needs to be One higher than that on the second card, so possibilities are : $\begin{array}{c} \begin{array}{cc} 1^{\text{st}} \text{ card} & 2^{\text{nd}} \text{ card}\\ \hline \color{red}1 & \color{red}-\\ 2 & 1\\ 3 & 2\\ 4 & 3\\ 5 & 4\\ \color{red}- & \color{red}5 \end{array}\\ \hline \text{Total $:4$ possibilities} \end{array}$ Total possible ways of picking up the cards $= 5 \times 4 = 20$ Thus, the required Probability $= \dfrac{\text{favorable ways}}{\text{total possible ways}}= \dfrac{4}{20} = \dfrac 15$ Option A is correct amarVashishth answered Oct 22, 2015 • selected Nov 25, 2015 by Pooja Palod amarVashishth comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments MANSI_SOMANI commented Dec 29, 2022 reply Follow Share The point which i missed :- they are asking only 1 higher 0 votes 0 votes MANSI_SOMANI commented Dec 29, 2022 reply Follow Share “Two cards are then removed one at a time from the deck” i think due to this line it’s w/o replacement M I right? @Umair alvi 0 votes 0 votes ꧁༒☬ĿọŗԀ 🆂🅷🅸🆅🅰☬༒꧂ commented Oct 19, 2023 reply Follow Share yes without replacement they are using . 0 votes 0 votes Please log in or register to add a comment.
26 votes 26 votes Here we should consider without replacement, since "removed one at a time" means the card has been removed from the deck. Prob of picking the first card = 1/5 Now there are 4 cards in the deck. Prob of picking the second card = 1/4 Possible favourable combinations = 2-1, 3-2, 4-3, 5-4 Probability of each combination = (1/5)*(1/4) = 1/20 Hence answer = 4*1/20 = 1/5 Dhananjay answered Jan 18, 2015 Dhananjay comment Share Follow See all 0 reply Please log in or register to add a comment.
7 votes 7 votes the probability of choosing first no's is =(2,3,4,5)/(1,2,3,4,5)=4/5 the second time, we have only one option to choose out of four option=1/4 so,the total probability=(4/5)*(1/4)=1/5 Vikash answered Nov 14, 2016 Vikash comment Share Follow See all 2 Comments See all 2 2 Comments reply Pradeep Kumar 5 commented Nov 25, 2018 reply Follow Share How to know total possible ways of picking a card. That is 5*4=20 0 votes 0 votes Verma Ashish commented Jan 9, 2019 reply Follow Share $^5P_2$ 0 votes 0 votes Please log in or register to add a comment.
5 votes 5 votes with 5 cards to choose we can only fulfil the condition if we pick 2,3,4,5 in our choice else theres no way to get the same so 4/(5*4) is the answer Bhagirathi answered Oct 10, 2014 Bhagirathi comment Share Follow See all 0 reply Please log in or register to add a comment.