Ironic problem

Probability Level 4

My English teacher at school once told me she didn't like teaching a lesson in a classroom that had had a Math lesson right before hers, because the board was always full of "things to erase", and these things make her upset, since, as she said, she's not very good at Math. Suppose a day of lessons in my school has six different lessons: there are no repeated subjects in the same day. Considering 12 different subjects, Math and English included, what is the probability of my English teacher don't have to teach a lesson right after a Math lesson in a random day of lessons in my school?

Details and Assumptions :

  • Take the lessons to be taught randomly: there is no defined order for lessons.

  • Exception : between the third and the fourth lesson, there is a break. After going to the break, the teacher that taught the third lesson always erases the board, so that my English teacher doesn't mind about what was the last lesson that was given if she teaches the fourth lesson.

  • The answer must be a number that represents the percentage of probability: if you reach the probability fraction, take the percentage and just use one decimal. For instance, 89 157 = 0.5668789809 = 56.68789809 \frac{89}{157} = 0.5668789809 = 56.68789809 %, so the answer would be 56.6 56.6 .


The answer is 46.1.

Bruno Oggioni Moura by Bruno Oggioni Moura , 21, Brazil

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1 solution

The English teacher can teach a lesson in any of the six "places" (the first lesson, the second lesson,..., the sixth lesson), but the lesson right before cannot be a Math one, unless the English teacher teaches the fourth lesson.

English lesson being the first one: p 1 = 1 12 × 11 11 × 10 10 × . . . × 7 7 = 1 12 p_{1} = \frac{1}{12} \times \frac{11}{11} \times \frac{10}{10} \times ... \times \frac{7}{7} = \frac{1}{12}

English lesson being the second one: p 2 = 10 12 × 1 11 × 10 10 × 9 9 × . . . × 7 7 = 10 132 p_{2} = \frac{10}{12} \times \frac{1}{11} \times \frac{10}{10} \times \frac{9}{9} \times ... \times \frac{7}{7} = \frac{10}{132}

English lesson being the third one: p 3 = 11 12 × 9 11 × 1 10 × 9 9 × 8 8 × 7 7 = 99 1320 p_{3} = \frac{11}{12} \times \frac{9}{11} \times \frac{1}{10} \times \frac{9}{9} \times \frac{8}{8} \times \frac{7}{7} = \frac{99}{1320}

English lesson being the fourth one: p 4 = 11 12 × 10 11 × 9 10 × 1 9 × 8 8 × 7 7 = 110 1320 p_{4} = \frac{11}{12} \times \frac{10}{11} \times \frac{9}{10} \times \frac{1}{9} \times \frac{8}{8} \times \frac{7}{7} = \frac{110}{1320}

English lesson being the fifth one: p 5 = 11 12 × 10 11 × 9 10 × 7 9 × 1 8 × 7 7 = 7 96 p_{5} = \frac{11}{12} \times \frac{10}{11} \times \frac{9}{10} \times \frac{7}{9} \times \frac{1}{8} \times \frac{7}{7} = \frac{7}{96}

English lesson being the sixth (last) one: p 6 = 11 12 × 10 11 × 9 10 × 8 9 × 6 8 × 1 7 = 6 84 p_{6} = \frac{11}{12} \times \frac{10}{11} \times \frac{9}{10} \times \frac{8}{9} \times \frac{6}{8} \times \frac{1}{7} = \frac{6}{84}

S = p 1 + p 2 + p 3 + p 4 + p 5 + p 6 = 5689 12320 = 0.4617694805 S = p_{1} + p_{2} + p_{3} + p_{4} + p_{5} + p_{6} = \frac{5689}{12320} = 0.4617694805 , hence the answer is 46.1 \boxed{46.1}

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