A deck of 5 cards each carrying a distinct number from 1 to 5 is shuffled thoroughly Two cards are

no of out comes=4[(2,1),(3,2),(4,3),(5,4)]
total no of out comes=5c2=10
probability=4/10=2/5
9 years agoHelpfull: Yes(24) No(10)

ans will be 1/5
probability of selecting two cards from 5 cards= 5p2= 20 ways
and 4 possibilities 1) (5comes before4),
2) ( 4 comes before 3)
3) (3 comes before 2)
4) (2 comes before 1)
Therefore 4/20=1/5
9 years agoHelpfull: Yes(17) No(11)
total ways are 20 not 10
1,2,3,4,5 are nos...on cards
ways:-
(1,2) (2,1) (1,3)(3,1)...and so on ....2,1 is a different case than 1,2 hence these are total 20 ways..
pallavi is correct ans is 1/5
8 years agoHelpfull: Yes(5) No(0)
Numbers on card 1,2,3,4,5

p{5}-[1/5 * 4/5]
p{4}-[1/5 * 3/5]
p{3}-[1/5 * 2/5]
p{2}-[1/5 * 1/5]
p{1}-[0/5 * 0/5]

Ans-p{5}+p{4}+p{3}+p{2}+p{1}

Ans- 1/25 {{4+3+2+1}}

Ans 2/5
9 years agoHelpfull: Yes(4) No(1)
Kuch v likh kr confuse kyo krte ho Kamino...!!

Ans will be : 2/5
P(A) = favourable event / Total outcomes.

Now, total outcomes = { (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) }
Total = 20

Note: we can't take (1,1) or (2,2).....(5,5) bcoz card are numbered 1 to 5 only
agar aapne 1 card drawn kiya let (2) den fir se 2 aana possible nhi h!!! Got it.

Now, favorable event [ choose from total events thar is differ by 1]
= { (1,2), (2,1) ,(2,3),(3,2), (3,4), (4,3) , (4,5), (5,4) }
= 8
P(A) = fav/total = 8/20 = 2/5 Ans.
8 years agoHelpfull: Yes(4) No(9)
You have to select 2 cards from 5.

Since the order in which they are drawn matters,
there are 5P2 = 5!/3! = 20 elementary events
from which there are 4 favorable number of cases:

5 before 4, 4 before 3, 3 before 2 and 2 before 1.

Hence, probability = 4/20 = 1/5
7 years agoHelpfull: Yes(3) No(0)
xs458xsdsd
9 years agoHelpfull: Yes(2) No(1)
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
8 years agoHelpfull: Yes(2) No(0)
ans ): 1 that is 1/5
8 years agoHelpfull: Yes(2) No(0)
totally confused people.
i think 1/5 will correct cause sample space is 20... and there are four favorable outcomes so 4/20=1/5
8 years agoHelpfull: Yes(2) No(0)
card contain number(n=1,2,3,4,5)
so two numbers can be taken in total 5c2=10 ways.
but condition is that first number must be one greater than second one so
E={(2,1),(3,2),(4,3),(5,4)}
n(E)=4
p(E)=n(E)/n(s)
=4/10
=2/5
option 4) is the right answer.
9 years agoHelpfull: Yes(1) No(3)
ANSWER IS 2/5
NO.OF WAYS TO SELECT ANY TWO OF CARDS FROM DECK OF 5 CARDS IS 25 WAYS

two cards are selected with the number on the first card being one higher than the number on the second card IS 10 WAYS
THEN THE PROBABILITY IS 10/25=2/5
9 years agoHelpfull: Yes(0) No(5)
when 5 card are suffled , the total event is 5!
drawing two card may be (2,1) (3,2) ( 3,1) (4,3)(4,2)(4,1)(5.4)(5,3)(5,2)(5,1)
10 cards can be drawn where first one is bigger than another

so 10/5! = 1/12
9 years agoHelpfull: Yes(0) No(6)
when we select the first card as 1 them the next card should be 2,3,4,5
and similarly when we select 2 nxt card should be 3,4,5
thus the probability would be
1/5*4/5+1/5*3/5+1/5*2/5+1/5*1/5
thus we get 10/25=2/5
8 years agoHelpfull: Yes(0) No(1)
no of possible outcome: (1,2)(1,3)(1,4)(1,5),(2,3)(2,4)(2,5)(3,4)(3,5)(4,5).
total no of possiblities: 5*5=25
req. probability: 10/25=2/5
8 years agoHelpfull: Yes(0) No(1)
deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at 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) 1/5
(B) 4/25
(C) 1/4
(D) 2/5

Answer: (A)

Explanation: You have to select 2 cards from 5.

Since the order in which they are drawn matters,
there are 5P2 = 5!/3! = 20 elementary events
from which there are 4 favorable number of cases:

5 before 4, 4 before 3, 3 before 2 and 2 before 1.

Hence, probability = 4/20 = 1/5
6 years agoHelpfull: Yes(0) No(0)
Bhai one at a time hai question is confusing there can or cannot be a replacement.no information is given regarding replacement.But logically if u think replacement must not be there because it is given
1st. 2 cards drawn one at a time
2nd then you compare viewing both the card
So there are 4 possible outcomes
Probability of each outcome is same
P×4
Now 1st 1 card is drawn it's probability is 1/5. AND
Now second card is drawn probability=1/4
Total p=1/5×1/4=1/20
Now probability of 4 outcomes =4×1/20=1/5
6 years agoHelpfull: Yes(0) No(0)
You have to select 2 cards from 5.

Since the order in which they are drawn matters,
there are 5P2 = 5!/3! = 20 elementary events
from which there are 4 favorable number of cases:

5 before 4, 4 before 3, 3 before 2 and 2 before 1.

Hence, probability = 4/20 = 1/5
6 years agoHelpfull: Yes(0) No(0)
100% Accurate
You have to select 2 cards from 5.

Since the order in which they are drawn matters,
there are 5P2 = 5!/3! = 20 elementary events
from which there are 4 favorable number of cases:

5 before 4, 4 before 3, 3 before 2 and 2 before 1.

Hence, probability = 4/20 = 1/5
6 years agoHelpfull: Yes(0) No(0)
You have to select 2 cards from 5.

Since the order in which they are drawn matters,
there are 5P2 = 5!/3! = 20 elementary events
from which there are 4 favorable number of cases:

5 before 4, 4 before 3, 3 before 2 and 2 before 1.

Hence, probability = 4/20 = 1/5
4 years agoHelpfull: Yes(0) No(0)
sequence of cards matters here. So we have to use permutation.. not combination.
so we arrange 2 cards among 5 cards by 5p2=20

and there are 4 favorable number of cases:

5 before 4, 4 before 3, 3 before 2 and 2 before 1
so =4/20=1/5 ans.
1 year agoHelpfull: Yes(0) No(0)

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