  Question

In a multiple-choice question, these are five alternative answers, of which one or more than one correct. A candidate will get marks on the question if he ticks all the correct answers. The candidate decides to tick answers at random. How many chances should he be allowed so that the probability of his getting marks on the question exceeds $$1/8\ ?$$

Solution

Total number of ways to answer the question $$^{ 5 }{ { C }_{ 1 } }+^{ 5 }{ { C }_{ 2 } }+^{ 5 }{ { C }_{ 3 } }+^{ 5 }{ { C }_{ 4 } }+^{ 5 }{ { C }_{ 5 } }={ 2 }^{ 5 }-1=31$$P(Correct answer in I chance)=$$\cfrac { 1 }{ 31 }$$P(Correct answer in II chance)=$$\cfrac { 30 }{ 31 } .\cfrac { 1 }{ 30 } =\cfrac { 1 }{ 31 }$$P(Correct answer in III chance)=$$\cfrac { 30 }{ 31 } .\cfrac { 29 }{ 30 } .\cfrac { 1 }{ 29 } =\cfrac { 1 }{ 31 }$$P(Correct answer in IV chance)=$$\cfrac { 30 }{ 31 } .\cfrac { 29 }{ 30 } .\cfrac { 28 }{ 29 } .\cfrac { 1 }{ 28 } =\cfrac { 1 }{ 31 }$$P(getting marks) exceeds $$\cfrac { 1 }{ 8 }$$ in $$4$$ chances.Mathematics

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