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655-705 Level|   Probability|      
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Bunuel
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In a class of 40 students, 12 are left-handed and the other 28 are rig 18 Dec 2019, 03:12
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55% (hard)

Question Stats:

61% (01:59) correct 39% (01:31) wrong based on 92 sessions

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In a class of 40 students, 12 are left-handed and the other 28 are right-handed. If two students are chosen at random, what’s the probability that one is left-handed and one is right-handed?

A. 21/100
B. 14/65
C. 21/50
D. 28/65
E. 37/65
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D
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In a class of 40 students, 12 are left-handed and the other 28 are rig Updated on: 19 Dec 2019, 13:20
Originally posted by ankitbansal89 on 18 Dec 2019, 14:25.
Last edited by ankitbansal89 on 19 Dec 2019, 13:20, edited 1 time in total.
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12/40*28/39+28/40*12/39
28/65 ans
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In a class of 40 students, 12 are left-handed and the other 28 are rig 19 Dec 2019, 07:32
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ankitbansal89
12/40*28/39+28/40*12/39
14/65 ans
total 40c2
favrable outcome 12c1 * 28c1

ans 28/65
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In a class of 40 students, 12 are left-handed and the other 28 are rig 19 Dec 2019, 10:06
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possible ways
2c1 * 12/40 * 28/39 = 28/65
IMO D

Bunuel
In a class of 40 students, 12 are left-handed and the other 28 are right-handed. If two students are chosen at random, what’s the probability that one is left-handed and one is right-handed?

A. 21/100
B. 14/65
C. 21/50
D. 28/65
E. 37/65
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In a class of 40 students, 12 are left-handed and the other 28 are rig 19 Dec 2019, 10:24
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\(\frac{(12C1*28C1)}{40C2}\) = \(\frac{12*28}{20*39}\) = 28/65

IMO, D

Bunuel
In a class of 40 students, 12 are left-handed and the other 28 are right-handed. If two students are chosen at random, what’s the probability that one is left-handed and one is right-handed?

A. 21/100
B. 14/65
C. 21/50
D. 28/65
E. 37/65
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In a class of 40 students, 12 are left-handed and the other 28 are rig 26 Dec 2019, 21:12
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Bunuel
In a class of 40 students, 12 are left-handed and the other 28 are right-handed. If two students are chosen at random, what’s the probability that one is left-handed and one is right-handed?

A. 21/100
B. 14/65
C. 21/50
D. 28/65
E. 37/65

(12C1 x 28C1)/(40C2) = (12 x 28)/[(40 x 39)/2] = (12 x 28)/(20 x 39) = (4 x 7)/(5 x 13) = 28/65

Alternate Solution:

We can choose either (R- L) or (L- R) to satisfy the requirement. The probability of (R-L) is 28/40 x 12/39 = 14/65. The probability of (L-R) is 12/40 x 28/39 = 14/65. Thus the total probability is 14/65 + 14/65 = 28/65.

Answer: D
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