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11 Sep 2026, 10:08
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
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Question Stats:
64% (02:34) correct
36%
(02:51)
wrong
based on 44
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An envelope contains ten bills: 3 ones, 3 fives, 2 tens, and 2 twenties. Three bills are drawn at random without replacement. What is the probability that at least $20 is drawn in total?
(A) 5/12
(B) 7/12
(C) 19/30
(D) 13/20
(E) 2/3
(A) 5/12
(B) 7/12
(C) 19/30
(D) 13/20
(E) 2/3
11 Sep 2026, 18:50
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Lets find out the probability of less than $20 amount. The cases for this would be with probability calculation:
1. 3 $1 bills = (3/10)(2/9)(1/8)
2. 3 $5 bills = (3/10)(2/9)(1/8)
3. 2 $1 and 1 $5 bills = (3/10)(2/9)(3/8) x 3
4. 1 $1 and 2 $5 bills = (3/10)(2/9)(3/8) x 3
5. 1 $10, 1 $5, 1 $1 bill = (2/10)(3/9)(3/8) x 6
6. 1 $10 and 2 $1 bills = (2/10)(3/9)(2/8) x 3
Totalling all these probabilities = 11/30
So, the probability for atleast $20 amount = 1-(11/30) = 19/30. Ans is C
1. 3 $1 bills = (3/10)(2/9)(1/8)
2. 3 $5 bills = (3/10)(2/9)(1/8)
3. 2 $1 and 1 $5 bills = (3/10)(2/9)(3/8) x 3
4. 1 $1 and 2 $5 bills = (3/10)(2/9)(3/8) x 3
5. 1 $10, 1 $5, 1 $1 bill = (2/10)(3/9)(3/8) x 6
6. 1 $10 and 2 $1 bills = (2/10)(3/9)(2/8) x 3
Totalling all these probabilities = 11/30
So, the probability for atleast $20 amount = 1-(11/30) = 19/30. Ans is C
11 Sep 2026, 20:30
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The solution above is valid , but has anybody found a faster approach ?
