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 18.0k 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.
2 votes 2 votes The possible events are (2,1) (3,2) (4,3) (5,4). So only 4 possibilities are there and sample space will be, 5C1 × 4C1 = 20 So probability = 4/20 = 1/5 Answer:A keshore muralidharan answered Sep 12, 2020 keshore muralidharan comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Number of ways cards can be arranged = 5! Favorable ways = 5 4 _ _ _ => 6 ways 4 3 _ _ _ => 6 ways 3 2 _ _ _ => 6 ways 2 1 _ _ _ => 6 ways (6+6+6+6)/5! => 1/5 => A arjuno answered Jan 11, 2020 arjuno comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes $\binom{1}{1} * \binom{4}{1}/\binom{5}{1}*\binom{4}{1}= 1/5$ Anit Pratap answered Aug 8, 2020 Anit Pratap comment Share Follow See all 0 reply Please log in or register to add a comment.