A table can help in finding some combined events. Suppose two coins are tossed at once. The results can be arranged as follows:

2nd Coin \ 1st Coin	H	T
H	HH	HT
T	TH	TT

Sample space S = {HH, HT, TH, TT}; n(S) = 4

Three coins are tossed at once. Find the probability of obtaining:

a) three tails b) one head and two tails c) at least two heads

Solution:

Let H represent heads and T represent tails. Using a tree diagram:

Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}; n(S) = 8

a) Event A (three tails): A = {TTT}, n(A) = 1 P(A) = n(A) / n(S) = 1/8

b) Event B (one head and two tails): B = {HTT, THT, TTH}, n(B) = 3 P(B) = n(B) / n(S) = 3/8

c) Event C (at least two heads): C = {HHH, HHT, HTH, THH}, n(C) = 4 P(C) = n(C) / n(S) = 4/8 = 1/2

For any combined events A and B from a sample space S,

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

If $P(A) = 1/4$, $P(B) = 1/3$, and $P(A \cup B) = 1/2$, find $P(A \cap B)$.

Solution:

P(A ∩ B) = P(A) + P(B) − P(A ∪ B) P(A ∩ B) = 1/4 + 1/3 − 1/2 P(A ∩ B) = (3 + 4 - 6)/12 P(A ∩ B) = 1/12

If a six-sided die is rolled, let B be the event that the number is even, and C be the event that the number is greater than 4. Find the probability that the number is even or greater than four.

Solution:

S = {1, 2, 3, 4, 5, 6}, n(S) = 6 B = {2, 4, 6}, n(B) = 3, P(B) = 3/6 = 1/2 C = {5, 6}, n(C) = 2, P(C) = 2/6 = 1/3 B ∩ C = {6}, n(B ∩ C) = 1, P(B ∩ C) = 1/6 P(B ∪ C) = P(B) + P(C) − P(B ∩ C) P(B ∪ C) = 3/6 + 2/6 − 1/6 P(B ∪ C) = 4/6 = 2/3

Given that $P(A) = 0.5$, $P(B) = 0.7$, and $P(A \cap B) = 0.3$. Using a Venn diagram, find:

(a) $P(A' \cup B')$ (b) $P(B \cap A')$ (c) $P(A \cup B')$

Solution:

(a)

P(A′ ∪ B′) = 1 − P(A ∩ B) = 1 − 0.3 = 0.7

Alternatively,

P(A′ ∪ B′) = 0.1 + 0.4 + 0.2 = 0.7

(b)

P(A ∪ B′) = 1 − P(B − A) = 1 − 0.4 = 0.6

Alternatively,

P(A ∪ B′) = 0.1 + 0.2 + 0.3 = 0.6

Remark: If P(A) represents the probability of event A occurring and P(A′) represents the probability of event A not occurring, then P(A) + P(A′) = 1