A box contains 10 cards numbered 1 through 10. If 3 cards are randomly

Question

A box contains 10 cards numbered 1 through 10. If 3 cards are randomly selected without replacement, what is the probability that the highest-numbered card picked is 5?

A. 3/10
B. 1/6
C. 1/10
D. 1/12
E. 1/20

Bunuel · Accepted Answer

Official Solution:

A box contains 10 cards numbered 1 through 10. If 3 cards are randomly selected without replacement, what is the probability that the highest-numbered card picked is 5?

A.  \frac{3}{10}

B.  \frac{1}{6}

C.  \frac{1}{10}

D.  \frac{1}{12}

E.  \frac{1}{20}

For the highest-numbered card selected to be 5, we should select card numbered 5, and any two of the cards numbered 1, 2, 3, and 4. The number of ways to select 5 is obviously 1, and the number of ways to select two cards from 4 cards (cards numbered 1, 2, 3, and 4) is  C^2_4 . The total number of ways to select three cards out of 10 is  C^3_{10} , therefore the probability is:

\frac{1*C^2_4}{C^3_{10}} = \frac{6}{120} = \frac{1}{20}

Answer: E

waytowharton · Answer

E. (1/20) 
Step 1 : 3 numbers would be 2 numbers between(1-4) and other one 5. Ways to select = 4c2*1
Step 2 : Combinations - 3!*4c2 = 36
Step 3 : Denominator - 10*9*8
Answer - 36/(10*9*8) = 1/20

Archit3110 · Answer

for 5 be the largest value among 3 cards the other two cards can be from 1 to 4
ways to select these two cards 4c2 ; 6 ways
total ways to choose 10c3 cards ; 120
P = 6/120 ; 1/20
option E

sujoykrdatta · Answer

We need to pick 3 cards with 5 as the largest number.
Let us pick 5 first.
We now need to pick any 2 numbers from 1 to 4, i.e. 4 numbers
Number of favourable ways = 4C2 = 6

Total ways of choosing any 3 numbers = 10C3 = 120

Probability = 6/120 = 1/20

Note: we didn't consider arrangement in either numerator or denominator. If we did, it would be:
Favourable cases = 4C2 x 3! = 36
Total cases = 10 x 9 x 8 = 720
Probability = 36/720 = 1/20

Answer E

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demawtf · Answer

A box contains 10 cards numbered 1 through 10. If 3 cards are randomly selected without replacement, what is the probability that the highest-numbered card picked is 5?

A. 3/10
B. 1/6
C. 1/10
D. 1/12
E. 1/20

Here the order doesn't matter because there is not an explicit connotation of 1st, 2nd or 3rd card and the activity is done without replacement, we are in the combinations scenario.

tot comb = 10!/7!*3! = 120
desired (1,2,5)(1,3,5)(1,4,5)(2,3,5)(2,4,5)(3,4,5) so 6 
or better we need to have 5 for sure and the other 2 numbers need to be picked out from 4 numbers (1 to 4) in order that the 5 is the highest --> so the desired can be calculated as 4!/2!*2!=6

desired/tot = 6/120 is 1/20 so Answer is E.

Cligzy · Answer

Hi people:

Might be a stupid question but I'll ask it anyways:

Why isn't it 4c1 + 3c1 in the numerator over 10c3. Heres my logic, we have to pick 1 one for 4 cards, and then we have to pick 1 of the 3 remaining cards after that?

Why doesn't my logic hold here?

Thanks in advance

Krunaal · Answer

1. You would want to do 4C1 * 3C1 (multiply, and not add) if you want to pick 1 of 4 first and 1 of 3 then.

2.  But in this case that's wrong too,  because when you do that you count the cases twice. Let's say you pick 1 first and then 2, in another case you're picking 2 first and then 1. In essence, for both cases you're picking {1,2}, it shouldn't matter what you pick first. Therefore, we directly use 4C2 and pick 2 cards from the remaining 4, the arrangement or order of selection doesn't matter. Hope it helps.

Cligzy · Answer

Thank for the quick replay and the good explanation

But why are we using combinations and not permutations. In this question, my intuition says it does.

Krunaal · Answer

We use permutations when the order of selection / arrangement matters. In this question we are just asked to pick 3 cards out of 10 - no matter in what order they are picked.

GMAT2point0 · Answer

Can someone solve this using the sequential probability method (not sure if that's the name)?

Probability of the first card being 5 = 1/10
Probability of second card being one from 1,2,3,4 = 4/9
Probability of third card being one from the remaining set of 1,2,3,4 = 3/8

Now the probability of this sequence is 1/10 * 4/9 * 3/8 = 1/60

I know we must multiply by this by X, to account for the fact that these cards can be given to us in other sequences, and X will be equal to 3 to get the final answer as 1/20, but I'm not sure how to logically get the value of X.

Please help! Thanks!

luvsic · Answer

its because the card being 5 can be any of the 3 cards picked (not just the 1st one), so you have to multiply by 3 to account for this

arunbhati · Answer

Hi  Bunuel ,

please explain 1/10*4/9*3/8*3! which is coming as 1/10 why this approach is wrong

Bunuel · Answer

The mistake is the 3!.

In 1/10 * 4/9 * 3/8 you are treating the draw as: card 5 first, then one card from 1 to 4, then another card from 1 to 4.

The two smaller cards are already being counted in order. For example, 2 then 4 and 4 then 2 are already separate possibilities.

So you do not need to arrange all 3 cards with 3!. You only need to account for the 3 possible places where card 5 could appear.

Correct multiplier is 3, not 3!:

1/10 * 4/9 * 3/8 * 3 = 1/20

So the approach overcounts by a factor of 2.

Or think about it this way: we need P(5, less than 5, less than 5) = 1/10 * 4/9 * 3/8 * 3!/2!. We multiply by 3!/2! because {5, less than 5, less than 5} can be arranged in 3!/2! = 3 ways. The two “less than 5” cards are of the same type in this setup, so we divide by 2!.

Hope it helps.

Adit_ · Answer

We need to select 2 numbers below 5 basically.
4C2 ways.
Once we have them selected, we can arrange the 3 numbers in 3! ways.

Thus 4C2*3! / 10*9*8 = 36/10*9*8 =  1/20 .

A box contains 10 cards numbered 1 through 10. If 3 cards are randomly

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