Question

Chapter 2.2 Questions l. You are having a party, and have the guests choose out of a box of goodies at randonm. There are 2 party hats (H), 3 yo-yos (Y), and 1 box of turtle chocolates (T) (a) Construct a tree diagram for two guests choosing 1 goody each from the bag. Give yourself lots of room to draw it. (b) What is the probability of choosing at least one party hat? (c) What is the probability of no guest picking the turtle chocolates? (d) What is the probability of choosing at least one yo-yo given that the guests picked at least one party hat? (e) What is the probability of choosing 2 yo-yos given that 1 box of turtle chocolates was chosen? (f) What is the probability of choosing at least one party hat given that no yo-yos were chosen?

Answer 1

(a)

(b)

From the Tree diagram, probability of choosing at least one party hat is,

P(G1 = H) + P(G1 = Y, G2 = H) + P(G1 = T, G2 = H)

= (2/6) + (3/6) * (2/5) + (1/6) * (2/5) = 0.6 = 3/5

(c)

From the Tree diagram, probability of no guest choosing the turtle chocolate is,

= 1 - probability of at least one guest choosing the turtle chocolate

= 1 - [P(G1 = H, G2 = T) + P(G1 = Y, G2 = T) + P(G1 = T) ]

= 1- [ (2/6) * (1/5) + (3/6) * (1/5) + (1/6) ] = 1 - (1/3) = 2/3

(d)

From the Tree diagram, probability of choosing at least one yo-yo given the guests picked at least one party hat

= P(choosing at least one yo-yo | guests picked at least one party hat)

= P(choosing at least one yo-yo and guests picked at least one party hat) / P(guests picked at least one party hat)

= [P(G1 = H, G2 = Y) + (G1 = Y, G2 = H) ] / 0.6 (from part (b))

= [(2/6) * (3/5) + (3/6) * (2/5)] / 0.6

= 0.4 / 0.6

= 2/3

(e)

If a guest has already chosen box of turtle chocolates, then both guests cannot choose yo-yos. So, the probability of choosing 2 yo-yos is 0.

(f)

From the Tree diagram, probability of choosing at least one party hat is,

P(G1 = H, G2 = H) + P(G1 = H, G2 = T) + P(G1 = T, G2 = H)

= (2/6) * (1/5) + (2/6) * (1/5) + (1/6) * (2/5) = 0.2 = 1/5