I have two fair spinners. The first has the numbers 1,2,3, and 4. The second has the numbers 2,3,4 and 5. I spin both spinners and add the scores. What is the probability that the total is odd?

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Explanation:

It helps to make a table of all possible outcomes. Refer to the diagram below. Along the top we have the outcomes 1,2,3,4. Along the left side we have 2,3,4,5. Each row & column header is then added to form the corresponding inner cell. For example, in the top left corner we have 1+2 = 3.

There are 4 outcomes of spinner A and 4 outcomes of spinner B. This means we have 4*4 = 16 outcomes overall. Out of those 16 sums, we have the odd results {3,5,7,9} show up 1+3+3+1 = 8 times.

The probability we want is therefore 8/16 = 1/2. This is expected since half of the values of {1,2,3,4} are odd, and the same applies to {2,3,4,5}. So we should expect half of the sums to be odd as well.

Put another way: for any row or column, you'll find that the pattern is odd/even/odd/even or even/odd/even/odd. This shows that 50% of any row or column represents an odd number. This idea extends out to the entire table overall.

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

13.33%

say 50 oranges, then 100 apples

80 are red, 20 are yellow

total fruit = 150

20chances out of 150 for yellow

20/150 = 13.33%