M39-79

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.  \frac{3}{10} 
B.  \frac{1}{6} 
C.  \frac{1}{10} 
D.  \frac{1}{12} 
E.  \frac{1}{20}

Bunuel · 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

thisisakr · Answer

I did not quite understand the solution. i used another method in which I selected 1/10 * 4/9 * 3/8 as answer based on the the fact that prob of selecting 5 out of 10 cards is 1/10 then we are left with 4 options 1,2,3,4 so 4/9 and then 3/8.

Can someone let me know why this solution is wrong (1/60) is the answer using this method

Can someone let me know why this solution is wrong (1/60) is the answer using this method

Bunuel · Answer

If you use this method, you must account for all the ways the highest-numbered card (5) can appear in the selection. The three possible cases are:

5, less than 5, less than 5 
 Less than 5, 5, less than 5 
 Less than 5, less than 5, 5

Each of these cases has the same probability, so the correct approach would be:

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

This matches the correct answer. Your method undercounts because it only considers one specific order instead of all three possible ways the cards could be selected.

AVMachine · Answer

Why is this approch not working:
P(5) = 1/10; first draw
P(5>) = 4/9; second draw
P(5>) = 3/8; third draw

Now three draws can be in 3! ways = 6

Ans (1/10)*(4/9)*(3/8) * 6 = 1/10

What am I missing here?

Bunuel · Answer

Check here: https://gmatclub.com/forum/m39-440985.html#p3523552 With this method you should multiply by 3 not by 6, because the number of ways to arrange is 3: 5, less than 5, less than 5 Less than 5, 5, less than 5 Less than 5, less than 5, 5

gemministorm · Answer

I misinterpreted the question, i took it as from the 3 selected highest should not be more than 5 and ended up calculating
5c3/10c3....

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