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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?

1. $\left(\dfrac{1}{5}\right)$
2. $\left(\dfrac{4}{25}\right)$
3. $\left(\dfrac{1}{4}\right)$
4. $\left(\dfrac{2}{5}\right)$

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+1

$(1,1)$ $(1,2)$  $(1,3)$ $(1,4)$ $(1,5)$

$(2,1)$ $(2,2)$  $(2,3)$ $(2,4)$ $(2,5)$

.           $(3,2)$

.                      $(4,3)$

$(5,1)$ $(5,2)$  $(5,3)$ $(5,4)$ $(5,5)$

$\dfrac{4}{^5C_2}$  $\times$

$\dfrac{4}{^5P_2}$  $\checkmark$

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Why is 5C2 wrong ? I mean isn't it the total number of ways of selecting the two cards from 5 and after which we can simply interpret the answer as 4/(5c2).

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

by Boss
selected
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How to solve the question if it would be only HIGHER and not ONE HIGHER?
+14

How to solve the question if it would be only HIGHER and not ONE HIGHER?

Possible combinations :-

 First card second card 5 4,3,2,1 4 3,2,1 3 2,1 2 1 1 none

$P(First\ card\ is\ HIGHER\ than\ second\ one\ )=\left [ \left ( \frac{1}{5} \times \frac{4}{4}\right ) + \left ( \frac{1}{5} \times \frac{3}{4}\right ) + \left ( \frac{1}{5} \times \frac{2}{4}\right ) + \left ( \frac{1}{5} \times \frac{1}{4}\right ) \right]\\ =\frac{1}{2}$

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Got it
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I have a basic doubt. It is just given that we have 5 cards from 1-5, so should we assume that they are from same suite? Since nothing mentioned about suites of card, we can also have a sample space as selecting card with no.1 from either of the suite thus making sample space as 4*4*4*4*4?

​​​​Thanks

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
by Active
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
by
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How to know total possible ways of picking a card.   That is  5*4=20
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$^5P_2$
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

by Boss

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

by