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November 20, 2014

November 20, 2014

Posted by **Marylu** on Wednesday, March 23, 2011 at 6:19pm.

basketball, 52 like football, 61 like baseball, 29 like

baseball and football, 39 like football and

basketball, 42 like basketball and baseball, 27 like

all three sports. If a student is chosen at random,

what is the probability that she/he likes no more

than one sport

- math -
**Reiny**, Wednesday, March 23, 2011 at 7:20pmA - basketball

B - Football

C - baseball

Numer(A or B or C) = number(A) + number(B) + number(C) - number(A and B) - number(A and C) - number(B and C) + number(A and B and C)

100 = 70 +52 +61 -39 -29 - 42 + 27

Data is valid.

make Venn diagrams, 3 intersecting circles , call them A, B, and C

put 27 in the intersection of all 3

39 like football and basketball, but 27 of those have already been counted, so put 12 in the intersection of those two outside the 27

42 like baseball and basketball, but 27 of those have already been counted, so put 15 in the region of A and C outside the 27.

29 like baseball and football but 27 are already counted, so put 2 in the intersection of B and C, ouside the 27

now look at A, I count up 54 so far as liking Basketball, but we were told that 70 like basketball, which means we have to put 16 in the region of A not intersected with any of the others.

Continue in this way to fill in the other parts.

Now count up all the numbers of A, B, and C that do not intersect with any of the others.

Put that number of 100 to get your probability.

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