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Official Solution:

A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected one by one from the deck without replacement, what is the probability that the numbers on the cards are in increasing order?

A. \(\frac{1}{60}\)
B. \(\frac{1}{30}\)
C. \(\frac{1}{20}\)
D. \(\frac{1}{6}\)
E. \(\frac{1}{3}\)


Any three cards can be drawn in \(3! = 6\) ways, and only one sequence from these 6 will be in ascending order. So, the probability is \(\frac{1}{6}\).

For example, 2, 5, and 6 can be drawn in 6 ways ({firs card, second card, third card}): {2, 5, 6}, {2, 6, 5}, {5, 2, 6}, {5, 6, 2}, {6, 2, 5} and {6, 5, 2}. Only one from these 6 is in ascending order: {2, 5, 6}.


Answer: D
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In order for the 3 cards to be in sequence, the 1st card drawn must be a 1, 2, 3 or 4, so the odds of drawing one of these cards is 4/6. Without replacement, the next card must be in sequence, so the odds for drawing the next card in sequence is 1/5. Without the replacement, the same rules apply for the next card, so the odds of drawing the next card in sequence is 1/4. Total odds = 4/6 * 1/5 * 1/4 = 4/120 = 1/30
Ans= B
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Alternative approach :

Let the three cards be a , b and c

So the question becomes
is a<b<c ?

Now, a can have any three of the operations < = > with b and b can have any two operations with c
So total options = 3*2
Favourable: a<b<c ... Only one outcome.
Probablity = 1/6
D

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Total number of possible arrangements of 3 cards = 6C3 *3!

Total number of ways in which 3 card are in the increasing order = 6C3 * 1

Probability = 1/3! = 1/6


Bunuel
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3
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Bunuel
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3


You can choose 3 cards from 6 cards and arrange them in 6P3 ways; i.e. 120 ways. (Permutations)

You can choose 3 cards from 6 cards and arrange them in one way, i.e. 6C3; i.e. 20 ways; that is any unique combination of three numbers will effectively be in increasing order.

Ans is 20/120 = 1/6 (Option D)
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nick1816
Total number of possible arrangements of 3 cards = 6C3 *3!

Total number of ways in which 3 card are in the increasing order = 6C3 * 1

Probability = 1/3! = 1/6


Bunuel
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3

Hi Nick1816,
Can you please explain how have you counted the number in the increasing order as 6C3. Though I got the same result by counting the possible arrangements but I want to understand the thought process. Please explain.

Vinay
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Hey
Suppose you choose 1, 2 and 3.

Total possible arrangements of 1, 2 and 3 are 3!= 6. [(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2) and (3,2,1)]

But there is only 1 possible arrangement out of those 6 arrangements in which you can put these numbers in increasing order, i.e (1,2,3), and this is true for any combination of 3 distinct numbers.


Hence, Total number of ways in which 3 card are in the increasing order = [number of ways to choose 3 numbers out of 6] * [possible arrangements of those 3 numbers]= 6C3 * 1

If you still have any doubt, you can ask.




vinny12

Hi Nick1816,
Can you please explain how have you counted the number in the increasing order as 6C3. Though I got the same result by counting the possible arrangements but I want to understand the thought process. Please explain.

Vinay
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