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InterviewSolution
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There are five houses A, B, C, D and E in a school. There are 23 students in a class, in which 4 students in house A, Sin house B, 5 in house C, 2 in house D and remaining students are in the house E. Out of these one students is selected for being monitor. The probability that the student chosen is not of the houses A, B and C is : (A) \(\frac { 4 }{ 23 }\)(B) \(\frac { 6 }{ 23 }\)(C) \(\frac { 8 }{ 23 }\)(D) \(\frac { 17 }{ 23 }\) |
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Answer» Answer is (B) \(\frac { 6 }{ 23 }\) The total number of students in the class = 23 The sum of the students of the houses A, B and C = 4 + 8 + 5 = 17 The number of all possible outcomes = 23 The number of favourable outcomes = 17 The probability that the student choosen from the houses A, B and C = \(\frac { 17 }{ 23 }\) The probability that the ‘student not’ from the houses if A, B and C. = 1 – \(\frac { 17 }{ 23 }\) = \(\frac { 6 }{ 23 }\) |
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