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28 Jan 2022, 07:28
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D
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Difficulty:
95%
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Question Stats:
38% (02:43) correct
62%
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wrong
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A deck of cards contains 10 cards numbered from 1 to 10. Janet draws 5 of thse cards at random. She shows them to Rob, who tells her she has drawn the cards numbered 10 and 8, but not the card numbered 9 nor the card numbered 7. What is the probability that the sum of the numbers on the cards Janet has drawn is less than the sum of those she has not drawn?
(A) 1/4
(B) 3/10
(C) 1/3
(D) 7/20
(E) none of these
(A) 1/4
(B) 3/10
(C) 1/3
(D) 7/20
(E) none of these
31 Jan 2022, 17:46
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Bunuel
Probability and Combinations:
We know among the 5 cards, we have 8 and 10, but not 7 nor 9. Thus the three unknown cards must be between 1 and 6, inclusive.
Sum of Drawn: 18 + 3 more cards.
Sum of not Drawn: 16 + 3 more cards.
The total sum of 10 cards is \(\frac{1 + 10}{2}*10 = 55\), thus we want the total sum of drawn be 27 or less. The 3 drawn cards cannot exceed 9.
Thus {1, 2, 3} to {1, 2, 6} is viable. {1, 3, 4} and {1, 3, 5} is viable. Finally, {2, 3, 4} will also work. That is 7 scenarios total.
There are \(6C3 = \frac{6*5*4}{3!} = 20\) scenarios in total, thus the probability is \(\frac{7}{20}\).
Answer: D
11 May 2026, 14:06
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drawn=10+8=18
not drawn=9+7=16 i.e. the drawn is 2 more than the non drawn in terms of summation.
now we are left with 1,2,3,4,5,6 (total sum=21) ;
out of which we have to select 3 digits in total 6C3=20 ways.
however with a restriction that including those 3 digits the summation of the drawn ones is less than the non drawn ones.
in other words the selected ones must sum together less than 10 i.e. 1,2,3 1,2,4 1,2,5 1,2,6 1,3,4 1,3,5 & 2,3,4 (so that the earlier gap of 2 is nullified & exceeded).
hence required probability is 7/20 D
not drawn=9+7=16 i.e. the drawn is 2 more than the non drawn in terms of summation.
now we are left with 1,2,3,4,5,6 (total sum=21) ;
out of which we have to select 3 digits in total 6C3=20 ways.
however with a restriction that including those 3 digits the summation of the drawn ones is less than the non drawn ones.
in other words the selected ones must sum together less than 10 i.e. 1,2,3 1,2,4 1,2,5 1,2,6 1,3,4 1,3,5 & 2,3,4 (so that the earlier gap of 2 is nullified & exceeded).
hence required probability is 7/20 D
