[Ezra Halleck]

A 21 sided die is rolled (icosahedron). What is the probability that the roll is
a) even? b) a multiple of 3? c) even and a multiple of 3?
d) even or a multiple of 3? [use results from parts a), b) & c)]

[Eric Adarkwah]

possible outcome=21
a. element of even numbers in a 21 sided die ( 2, 4 6 8 10 12 14 16 18 20)
p( even) = 10/21 0r 0.47

multiples of 3 in 21 sided die ( 3 6 9 12 15 18 21)
P(multiples of 3)= 7/21 =1/3 or.33
c. P ( even and a multiple of 3) = ( 6 12 18) = 3/21 = 1/7 or .14
d. P( even or a multiply of 3)= 10/21+7/21-2(3/21)= 11/21 or .52

[Chana Max, RN]

the formula for d) is P(a)+P(b) – P(a intersect b) so10/21+7/21-3/21=14/21 or 0.67

[Ezra Halleck]

Yes, I agree with cmax. There is no “2” in the formula. By the way, I lied to you about the icosahedron. It actually has 20 sides, but let us pretend such a 21 sided die actually exists.

Finally, an interesting question is whether the 2 events in question are independent. To answer that we see if P(even)*P(mult of 3)=P(even and mult of 3).
10/21*1/3=?1/7

The answer is no (you can verify by using a calculator)

To see why without a calculator, notice that if the first fraction had been 9/21 instead of 10/21 then the product would have worked out to the RHS. Therefore, the LHS is in fact greater than the RHS and hence not the same.