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

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 .

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yes bcoz we have to take the ordered pairs in SS instead of unordered pairs here, order matters here.

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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 1^st card and 4 ways to choose the 2^nd card ,hence the |SS|=5x4=20

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

The point which i missed :- they are asking only 1 higher

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“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 .

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