It might seem that you could use the formula for the probability of two independent events and simply multiply [latex]\frac{4}{52}\cdot\frac{4}{52}=\frac{1}{169}[/latex]. This would be incorrect, however, because the two events are not independent. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck.

Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the conditional probability of drawing an ace. In this case the “condition” is that the first card is an ace. Symbolically, we write this as:

P(ace on second draw | an ace on the first draw).

The vertical bar “|” is read as “given,” so the above expression is short for “The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw.” What is this probability? After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is [latex]\frac{3}{51}=\frac{1}{17}[/latex].

Thus, the probability of both cards being aces is [latex]\frac{4}{52}\cdot\frac{3}{51}=\frac{12}{2652}=\frac{1}{221}[/latex].

The probability that the first card is a spade is [latex]\frac{13}{52}[/latex].

The probability that the second card is a spade, given the first was a spade, is [latex]\frac{12}{51}[/latex], since there is one less spade in the deck, and one less total cards.

The probability that both cards are spades is [latex]\frac{13}{52}\cdot\frac{12}{51}=\frac{156}{2652}\approx0.0588[/latex]

1. Since we know the person has a red car, we are only considering the 150 people in the first row of the table. Of those, 15 have a speeding ticket, so P(ticket | red car) = [latex]\frac{15}{150}=\frac{1}{10}=0.1[/latex]
2. Since we know the person has a speeding ticket, we are only considering the 60 people in the first column of the table. Of those, 15 have a red car, so P(red car | ticket) = [latex]\frac{15}{60}=\frac{1}{4}=0.25[/latex].

Notice from the last example that P(B | A) is not equal to P(A | B).

You can satisfy this condition by having Case A or Case B, as follows:

Case A) you can get the Ace of Diamonds first and then a black card or

Case B) you can get a black card first and then the Ace of Diamonds.

Let’s calculate the probability of Case A. The probability that the first card is the Ace of Diamonds is [latex]\frac{1}{52}[/latex]. The probability that the second card is black given that the first card is the Ace of Diamonds is [latex]\frac{26}{51}[/latex] because 26 of the remaining 51 cards are black. The probability is therefore [latex]\frac{1}{52}\cdot\frac{26}{51}=\frac{1}{102}[/latex].

Now for Case B: the probability that the first card is black is [latex]\frac{26}{52}=\frac{1}{2}[/latex]. The probability that the second card is the Ace of Diamonds given that the first card is black is [latex]\frac{1}{51}[/latex]. The probability of Case B is therefore [latex]\frac{1}{2}\cdot\frac{1}{51}=\frac{1}{102}[/latex], the same as the probability of Case 1.

Recall that the probability of A or B is P(A) + P(B) – P(A and B). In this problem, P(A and B) = 0 since the first card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is [latex]\frac{1}{101}+\frac{1}{101}=\frac{2}{101}[/latex]. The probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck is [latex]\frac{2}{101}[/latex].

1. Since we know the test result was positive, we’re limited to the 75 women in the first column, of which 5 were not pregnant. P(not pregnant | positive test result) = [latex]\frac{5}{75}\approx0.067[/latex].
2. Since we know the woman is not pregnant, we are limited to the 19 women in the second row, of which 5 had a positive test. P(positive test result | not pregnant) = [latex]\frac{5}{19}\approx0.263[/latex]

The second result is what is usually called a false positive: A positive result when the woman is not actually pregnant.