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# In a class of 40 students, 12 are left-handed and the other 28 are rig

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Re: In a class of 40 students, 12 are left-handed and the other 28 are rig [#permalink]
possible ways
2c1 * 12/40 * 28/39 = 28/65
IMO D

Bunuel wrote:
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 [#permalink]
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$$\frac{(12C1*28C1)}{40C2}$$ = $$\frac{12*28}{20*39}$$ = 28/65

IMO, D

Bunuel wrote:
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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Re: In a class of 40 students, 12 are left-handed and the other 28 are rig [#permalink]
Bunuel wrote:
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.