GATE CSE 2011 | Question: 34

Question

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?
 

• A. $\left(\dfrac{1}{5}\right)$  
• B. $\left(\dfrac{4}{25}\right)$  
• C. $\left(\dfrac{1}{4}\right)$  
• D. $\left(\dfrac{2}{5}\right)$

Comments

5--------1,2,3,4
4--------1,2,3
3--------1,2
2--------1

i too did same mistake :)

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This is a classic ordered sampling without replacement probability problem...Here order Matter.. so when order matter then we use "Permutation" not combination ... for choosing first card we have 5 choices after choosing first card remove it from the deck and we have 4 choices to choose the card ..so total number of possiblities = 5x4 =20 i.e 5P2

The sample space is

\begin{aligned} S=\{ &(1,2),(1,3),(1,4),(1,5),\\ &(2,1),(2,3),(2,4),(2,5),\\ &(3,1),(3,2),(3,4),(3,5),\\ &(4,1),(4,2),(4,3),(4,5),\\ &(5,1),(5,2),(5,3),(5,4) \} \end{aligned}

Favourable outcome will will be (first card , second card) where first = second+1

Number on the first card is exactly one higher than the number on the second card.

\begin{aligned} Favourable=\{ &(5,4),(4,3),(3,2),(2,1) \} \end{aligned}

Hence Prob = 4/20

 

 

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                              Start
                                |
             ┌──────────┬───────┼
             ↓          ↓       ↓       ↓          ↓
             1          2       3       4          5
            1/5        1/5     1/5     1/5        1/5
             |          |       |       |          |
             |          ↓       ↓       ↓          |
             |          1       2       3          |
             |         1/4     1/4     1/4         |
             |          |       |       |          |
             ✗         ✓       ✓       ✓          ✓
                       |       |       |          |
                     (2,1)   (3,2)   (4,3)      (5,4)

Answer 1

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

Comments

“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

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yes without replacement they are using .

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here most intresting and error pron point is  x no is ONE HIGHER  then y meance if it is 2 then second no is 1

if it is 3 then second no is 2,  if it is 4 then second no is 3 if it is 5 then second no is 4 like there will be 4 case

Answer 2

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

Answer 3

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

Comments

How to know total possible ways of picking a card.   That is  5*4=20

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$^5P_2$

Answer 4

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

Answer 5

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

Answer 6

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