maths , physics and chemistry books are stored on a library shelf that can accommodate 25 books . currently , 20 % of the shelf spots remain empty . there are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books . among all the books , 12 books are soft cover and the remaining are hard - cover . if there are a total of 7 hard - cover books among the maths and physics books . what is the probability q , that a book selected at random is either a hard cover book or a chemistry book ?
first phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf . to do so , you have the equations : m + p + c = 20 ( since 4 / 5 of the 25 spots are full of books ) m = 2 p p = 4 + c from that , you can use substitution to get everything down to one variable . c = p - 4 m = 2 p p = p then ( p - 4 ) + 2 p + p = 20 , so 4 p = 24 and p = 6 . that means that there are 12 math , 6 physics , and 2 chemistry books on the shelf . with those numbers , you also know that there are 8 total hardcovers , 1 of which is chemistry . so if your goal is to get either a hardcover or a chemistry , there are 9 ways towin - either one of the 7 hardcovers that are n ' t chemistry or the two chemistry books . so out of the 20 total , q = 9 provide the desired outcome , making the answer e .
subtract(add(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12))), multiply(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12))))