Probability of Combined Events practice questions

• Total number of outcomes $= 6 \times 8 = 48$
• Build the $6 \times 8$ grid of sums. The sums run from $2$ to $14$.
• Prime numbers in that range: $\{2, 3, 5, 7, 11, 13\}$
• Favourable outcomes: $(1,1)$, $(1,2)$, $(1,4)$, $(1,6)$, $(2,1)$, $(2,3)$, $(2,5)$, $(3,2)$, $(3,4)$, $(3,8)$, $(4,1)$, $(4,3)$, $(4,7)$, $(5,2)$, $(5,6)$, $(5,8)$, $(6,1)$, $(6,5)$, $(6,7)$
• Number of favourable outcomes $= 19$
• $P(\text{sum is prime}) = \frac{19}{48}$

Answer: $19/48$

• Tree diagram branches: First draw, Green (probability $\frac{2}{3}$) or Yellow (probability $\frac{1}{3}$). Since the bead is replaced, second draw has the same probabilities.
• $P(\text{green then yellow}) = \frac{2}{3} \times \frac{1}{3}$
• $= \frac{2}{9}$
• $P(\text{yellow then green}) = \frac{1}{3} \times \frac{2}{3}$
• $= \frac{2}{9}$
• $P(\text{one of each}) = P(GY) + P(YG)$
• $= \frac{2}{9} + \frac{2}{9}$
• $= \frac{4}{9}$

Answer: $4/9$

• There are $4$ Aces and $13$ clubs.
• The Ace of Clubs is both, so adding $4$ and $13$ would count it twice.
• Aces but not clubs: $4 - 1 = 3$.
• Clubs but not Aces: $13 - 1 = 12$.
• Cards that are Aces or clubs: $3 + 1 + 12 = 16$.
• $P(\text{an Ace or a club}) = \frac{16}{52}$
• $= \frac{4}{13}$

Answer: $4/13$