- This topic has 3 replies, 3 voices, and was last updated 8 years, 6 months ago by .
You must be logged in to reply to this topic.
Viewing 4 posts - 1 through 4 (of 4 total)
You must be logged in to reply to this topic.
You must be logged in to reply to this topic.
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)]
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
the formula for d) is P(a)+P(b) – P(a intersect b) so10/21+7/21-3/21=14/21 or 0.67
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.
You must be logged in to reply to this topic.
Ursula C. Schwerin Library
New York City College of Technology, C.U.N.Y
300 Jay Street, Library Building - 4th Floor
Our goal is to make the OpenLab accessible for all users.