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Probability

Example 2.7. Compute the probability of drawing from a pack of cards one that
is an ace or is a spade or shows an even number (2, 4, 6, 8, 10). ,

Solution. If, as previously, A is the event that an ace is drawn, P(A) = 52. Similarly the event
B, that a spade is drawn, has P(B) = 52. The further possibility C, that the card is even (but
not a picture card) has P(C) = 2052. The two-fold intersections have probabilities
1 5
P(A ∩ B) = , P(A ∩ C) = 0, P(B ∩ C) = .
52 52
There is no three-fold intersection as events A and C are mutually exclusive. Hence

1 31
P(A ∪ B ∪ C) = [(4 + 13 + 20) − (1 + 0 + 5) + (0)] = .
52 52
The reader should identify the 31 cards involved.
When the probabilities are combined to compute the probability for the union of the n general
events, the result, which may be proved by induction upon n, is
∑ ∑ ∑
P(A 1 ∪ A 2 ∪ . . . ∪ A n ) = P(A i ) − P(A i ∩ A j ) + P(A i ∩ A j ∩ A k )−
i i,j i,j,k
− . . . + (−1) n+1 P(A 1 ∩ A 2 ∩ . . . ∩ A n ). (2.14)

Each summation runs over all possible sets of subscripts, except those in which any two subscripts
in a set are the same. Equation (2.6) is a special case of (2.14) in which n = 2 and only the first two
terms on the RHS survive. We now illustrate this result with a worked example that has n = 4 and
includes a four-fold intersection.

Example 2.8. Find the probability of drawing from a pack a card that has at least
one of the following properties:

• A, it is an ace;

• B, it is a spade;

• C, it is a black honour card (ace, king, queen, jack or 10);

• D, it is a black ace. ,

Solution. Measuring all probabilities in units of 1 , the single-event probabilities are
52
P(A) = 4, P(B) = 13, P(C) = 10, P(D) = 2.

The two-fold intersection probabilities, measured in the same units, are
P(A ∩ B) = 1, P(A ∩ C) = 2, P(A ∩ D) = 2,

P(B ∩ C) = 5, P(B ∩ D) = 1, P(C ∩ D) = 2.

The three-fold intersections have probabilities
P(A ∩ B ∩ C) = 1, P(A ∩ B ∩ D) = 1, P(A ∩ C ∩ D) = 2, P(B ∩ C ∩ D) = 1.

Finally, the four-fold intersection, requiring all four conditions to hold, is satisfied only by the
ace of spades, and hence (again in units of 1 )
52
P(A ∩ B ∩ C ∩ D) = 1.
Substituting in (2.14) gives

1 30
P = [(4 + 13 + 10 + 2) − (1 + 2 + 2 + 5 + 1 + 2) + (1 + 1 + 2 + 1) − (1)] = .
52 52

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