A coin is tossed three times. Let A = [three heads occur] and B= [at least one head occurs]. What is the P(A Union B)? Please show how to solve ( and please solve without using permutation or combination).
P(A union B) = P(A) + P(B) - P(A and B)
=(1/2)^3 + (1-(1/2)^3) - (1/2)^3
=7/8
^A very similar solution is to note that A ⊆ B (if A occurs, then B necessarily occurs), so A ∪ B = B. Then P(A ∪ B) = P(B) = 1 - (1/2)^3 = 7/8.
just testing latex. i agree with @2015bl post.
[tex] \frac{3}{4} [/tex]
@MITer94 how did you use those symbols? through latex? what are the tags?
@iamjack not sure if this forum supports LaTeX input.
I did it the cheap way by opening a MS Word document, going into equation editor, and typing the \subseteq, \cup symbols there and pasting it here.