Exercise 13.2 · Q15: One card: in which cases are E and F independent?

Basics you need

Part (i): a spade, and an ace

1. Count and find the probabilities

   There are 13 spades and 4 aces, and exactly one card, the ace of spades, is both:

   $$\begin{aligned} P(E) &= \frac{13}{52} = \frac{1}{4}, \\ P(F) &= \frac{4}{52} = \frac{1}{13}, \\ P(E\cap F) &= \frac{1}{52}. \end{aligned}$$
   All 52 cards are equally likely, so each probability is a count over 52. The intersection "a spade and an ace" is a single card.
   $P(E) = \frac{n(E)}{n(S)}$⁠. Basics: classical probability

2. Apply the test

   $$\begin{aligned} P(E)\,P(F) &= \frac{1}{4}\times\frac{1}{13} \\ &= \frac{1}{52} = P(E\cap F). \end{aligned}$$

   So $E$ and $F$ are independent. Equivalently, $P(E|F) = \frac{1/52}{1/13} = \frac14 = P(E)$⁠: one ace in four is a spade, the same proportion as in the whole pack.

   The test holds exactly, so by Definition 3 the events are independent. This is the very example the book uses to introduce independent events at the start of Section 13.4.
   Definitions 2 and 3, Section 13.4. Lesson: independent events

Part (ii): a black card, and a king

1. Count and find the probabilities

   There are 26 black cards and 4 kings; the black kings are the kings of spades and clubs, 2 cards:

   $$\begin{aligned} P(E) &= \frac{26}{52} = \frac{1}{2}, \\ P(F) &= \frac{4}{52} = \frac{1}{13}, \\ P(E\cap F) &= \frac{2}{52} = \frac{1}{26}. \end{aligned}$$
   Black cards are the 13 spades and the 13 clubs. Of the 4 kings, two are in black suits and two in red suits.
   $P(E) = \frac{n(E)}{n(S)}$⁠. Basics: classical probability

2. Apply the test

   $$\begin{aligned} P(E)\,P(F) &= \frac{1}{2}\times\frac{1}{13} \\ &= \frac{1}{26} = P(E\cap F). \end{aligned}$$

   So $E$ and $F$ are independent. Equivalently, $P(E|F) = \frac{1/26}{1/13} = \frac12 = P(E)$⁠: half the kings are black, as is half the pack.

   Colour depends only on the suit and "king" only on the rank, and every rank appears once in every suit. So learning the card is a king does not change the chance it is black.
   Definition 3, Section 13.4. Lesson: independent events

Part (iii): king or queen, and queen or jack

1. Count and find the probabilities

   $E$ is the 4 kings and 4 queens, 8 cards; $F$ is the 4 queens and 4 jacks, 8 cards. A card in both must be a queen, so $E\cap F$ is the 4 queens:

   $$\begin{aligned} P(E) &= \frac{8}{52} = \frac{2}{13}, \\ P(F) &= \frac{8}{52} = \frac{2}{13}, \\ P(E\cap F) &= \frac{4}{52} = \frac{1}{13}. \end{aligned}$$
   A card cannot be a king and a jack at once, so the only cards in both events are the queens.
   "And" is an intersection. Basics: events as sets

2. Apply the test

   $$\begin{aligned} P(E)\,P(F) &= \frac{2}{13}\times\frac{2}{13} \\ &= \frac{4}{169}. \end{aligned}$$

   But $P(E\cap F) = \frac{1}{13} = \frac{13}{169}$⁠, and $\frac{13}{169} \ne \frac{4}{169}$⁠. So

   $$P(E\cap F) \ne P(E)\,P(F),$$

   and $E$ and $F$ are not independent.

   Writing both fractions over 169 makes the comparison plain. In conditional terms, $P(E|F) = \frac{1/13}{2/13} = \frac12$⁠, while $P(E) = \frac{2}{13}$⁠: learning that the card is a queen or jack raises the chance of "king or queen" from $\frac{2}{13}$ to $\frac12$⁠, because half of $F$ is queens.
   Definition 3 and Remark (i), Section 13.4. Lesson: independent events

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