M37-38

Question

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. \(\frac{1}{3}\)

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

Bunuel · Answer

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. \(\frac{1}{3}\)

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

Billyneutron · Answer

hello, can anyone explain why selecting 3 out of 6 is only 3!, shouldn't that be  /

Bunuel · Answer

This refers not to the number of ways to pick 3 cards out of 6, but rather to the order in which the three cards can be picked. For example, cards numbered 2, 3, and 6 can be picked in the following 6 orders:

2, 3, 6
2, 6, 3
3, 2, 6
3, 6, 2
6, 2, 3
6, 3, 2

As mentioned in the solution, only one of these six orders is in ascending order: 2, 3, 6.

mitshivi2024 · Answer

I had an alternate way of thinking :

1. We can select 3 cards from the deck in 6C3 ways and they can be arranged in => 6C3 *3! ways. 
2. Now out of these selections, only one way of each set would be unique, as in ascending order => 6C3

so , probability will be 6C3 / (6C3 *3!) = 1/3! = 1/6

I hope this makes sense.  Bunuel  can you please guide, if this line of thinking is correct?

Thanks in advance!

hr1212 · Answer

One way to think could be,

As cards are selected without replacement, total possible combinations = 6*5*4

No. of ways to select 3 cards out of 6 = 6C3 is also the number of ways these 3 cards can be picked in ascending order ie. let's say if you picked (4,2,5) as one of the combination then there is only one way (2,4,5) in which you can arrange them in an ascending order. So in a nutshell it's just like picking any 3 cards out of 6 and putting them in an ascending order.

Probability = Required/Total =  \frac{6C3}{6*5*4}  =  \frac{5*4}{6*5*4}  =  \frac{1}{6}

OmerKor · Answer

I did not quite understand the question. Hi

I had doubts regarding the phrase "increasing order". While it is possible to conclude that 2,4,5 is adequate, it is also possible to conclude that *only* 2,3,4/3,4,5 is correct. I believe it would be better to change the text slightly to make it more understandable. thanks!

Bunuel · Answer

"Increasing order" means arranging numbers from smallest to largest. It does not require the numbers to be consecutive. The wording is perfectly fine as it is.

siddharth_ · Answer

Total  arrangements  = selecting 3 out of 6 cards and then arranging every triplet = 6C3 * 3! OR 6P3 = 20*6
Favorable arrangements = 1 in each triplet * total triplets = 1*20
P = Favorable / Total = 20/(20*6) = 1/6

M37-38

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