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Updated on: 27 Apr 2018, 06:05
Originally posted by EgmatQuantExpert on 26 Apr 2018, 05:33.
Last edited by EgmatQuantExpert on 27 Apr 2018, 06:05, edited 2 times in total.
Last edited by EgmatQuantExpert on 27 Apr 2018, 06:05, edited 2 times in total.
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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:
25%
(medium)
Question Stats:
82% (02:14) correct
18%
(02:40)
wrong
based on 423
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3 deadly mistakes you must avoid in Probability - Exercise Question #3
3- Mark, Peter and David are trying to solve a probability problem. If the probability that Mark, Peter and David can solve the problem correctly is 1/5, 3/7, and 2/6 respectively.
What is the probability that exactly one student cannot solve the problem.
Options
To solve question 4: Question 4
To read the article: 3 deadly mistakes you must avoid in Probability
3- Mark, Peter and David are trying to solve a probability problem. If the probability that Mark, Peter and David can solve the problem correctly is 1/5, 3/7, and 2/6 respectively.
What is the probability that exactly one student cannot solve the problem.
Options
- a) 10/105
b) 16/105
c) 22/105
d) 27/105
e) 30/105
To solve question 4: Question 4
To read the article: 3 deadly mistakes you must avoid in Probability
Updated on: 29 Apr 2018, 02:47
Originally posted by EgmatQuantExpert on 26 Apr 2018, 05:34.
Last edited by EgmatQuantExpert on 29 Apr 2018, 02:47, edited 1 time in total.
Last edited by EgmatQuantExpert on 29 Apr 2018, 02:47, edited 1 time in total.
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Solution
Given:
- • Mark, Peter, and David are solving a probability problem.
• The probability that Mark will solve the problem correctly = \(\frac{1}{5}\)
• The probability that Peter will solve the problem correctly = \(\frac{3}{7}\)
• The probability that David will solve the problem correctly = \(\frac{2}{6}\)= \(\frac{1}{3}\)
To find:
•We need to find the probability such that exactly one person cannot solve the problem.
Approach and Working:
The possible cases such that exactly one cannot solve the problem are:
- • Mark and Peter can solve the problem while David cannot, Or,
• Mark and David can solve the problem while Peter cannot, Or,
• Peter and David can solve the problem while Mark cannot.
Thus, P(Exactly one cannot solve the problem) = P(M solves correctly)* P(P solves correctly)* P( D cannot)+ P(M solves correctly)* P(D solves correctly)* P( P cannot)+ P(P solves correctly)* P(D solves correctly)* P( M cannot)
- • P(Exactly one cannot solve the problem)= \(\frac{1}{5}\)*\(\frac{3}{7}\)*\(\frac{2}{3}\) + \(\frac{1}{5}\)* \(\frac{4}{7}\)*\(\frac{1}{3}\)+\(\frac{4}{5}\)*\(\frac{3}{7}\)*\(\frac{1}{3}\)
• = \(\frac{6}{105}+ \frac{4}{105}+ \frac{12}{105}\)
• =\(\frac{22}{105}\)
Answer: C
26 Apr 2018, 07:08
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EgmatQuantExpert
The probability that Mark and Peter can solve the problem correctly is \(\frac{1}{5}*\frac{3}{7} = \frac{3}{35}\)
The probability that Mark and David can solve the problem correctly is \(\frac{1}{5}*\frac{2}{6} = \frac{1}{15}\)
The probability that David and Peter can solve the problem correctly is \(\frac{3}{7}*\frac{3}{7} = \frac{1}{7}\)
The probability that Mark, David, and Peter can solve the problem correctly is \(\frac{1}{5}*\frac{3}{7}*\frac{2}{6} = \frac{1}{35}\)
Therefore, the probability that that exactly one student can't solve the problem is \(\frac{3}{35} + \frac{1}{15} + \frac{1}{7} - 3*(\frac{1}{35}) = \frac{7+15}{105} = \frac{22}{105}\) (Option C)
