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22 Jan 2020, 02:15
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Difficulty:
85%
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Question Stats:
55% (02:51) correct
45%
(02:29)
wrong
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Two students A and B participate in Physics and Chemistry exams. Each exam subject has 6 different exam codes and two exam subjects have their own different exam codes. In each exam subject, each student receives randomly exam code. What is the probability that A and B receive the same exam code in only one subject?
A. 7/18
B. 4/9
C. 2/9
D. 5/18
E. 1/36
Are You Up For the Challenge: 700 Level Questions
A. 7/18
B. 4/9
C. 2/9
D. 5/18
E. 1/36
Are You Up For the Challenge: 700 Level Questions
22 Jan 2020, 02:19
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Bunuel
Two students A and B participate in Physics and Chemistry exams. Each exam subject has 6 different exam codes and two exam subjects have their own different exam codes. In each exam subject, each student receives randomly exam code. What is the probability that A and B receive the same exam code in only one subject?
A. 7/18
B. 4/9
C. 2/9
D. 5/18
E. 1/36
Are You Up For the Challenge: 700 Level Questions
A. 7/18
B. 4/9
C. 2/9
D. 5/18
E. 1/36
Are You Up For the Challenge: 700 Level Questions
Total Combinations of exam codes = 6 (choices of Physics exam code)*6 (choices of Physics exam code) = 36
Required Probability = (36/36)*(1/36) = 1/36
Answer: Option E
22 Jan 2020, 03:27
Kudos
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Bunuel
Two students A and B participate in Physics and Chemistry exams. Each exam subject has 6 different exam codes and two exam subjects have their own different exam codes. In each exam subject, each student receives randomly exam code. What is the probability that A and B receive the same exam code in only one subject?
A. 7/18
B. 4/9
C. 2/9
D. 5/18
E. 1/36
A. 7/18
B. 4/9
C. 2/9
D. 5/18
E. 1/36
Required Probability = (same code in Physics & different code in Chemistry) or (same code in Chemistry & different code in Physics)
Probability of both getting same code in physics = Probability of both getting same code in chemistry = \(\frac{6}{6}*\frac{1}{6} = \frac{1}{6}\) [1st person can get from any 6 codes & 2nd person has to get exact code]
Probability of both getting different code in physics = Probability of both getting different code in chemistry = \(\frac{6}{6}*\frac{5}{6} = \frac{5}{6}\) [1st person can get from any 6 codes & 2nd person has to get from remaining 5 codes]
--> Required Probability = \(\frac{1}{6}*\frac{5}{6} + \frac{1}{6}*\frac{5}{6} = \frac{5}{36} + \frac{5}{36} = \frac{10}{36} = \frac{5}{18}\)
Option D
