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The Multiplication Rules Worksheet

Scenario #1: If we flip a coin, and it comes up heads, does that give us any information about the outcome of the next flip?

Scenario #2: Suppose we ask two random people in the U.S.the question: "Do you prefer dogs or cats?" Would knowing that first person answered "cats" give us any information about how the second person would answer?

Two events are independent if knowing that one event occurred doesn't change the probability that the other occurred.

Independent versus Not Independent

Example: Toss a coin. Knowing that heads came up on the first toss change the probability of heads coming up on the second toss. These two events are independent.

Example: Draw two cards from a standard, well-shuffled deck without replacement. Knowing that an ace was drawn first changes the probability drawing a second ace. These two events are NOT independent.

Example: On the other hand two cards from a standard, well-shuffled deck with replacement. Knowing that an ace was drawn first does not change the probability drawing another ace. These two events are independent.

Sampling With and Without Replacement

When we repeatedly sample from a population with replacement, each individual outcome is independent of the other. For example if we draw a card from a deck, put the card back, shuffle again, draw another card, each draw is independent.

On the other hand, if we sample repeatedly without replacement, the individual outcomes are NOT independent. For example, if we draw a card and do not put it back in the deck, the next outcome is influenced by the previous.

Definition: Two events $A$ and $B$ are independent if $$P(A|B)=P(A).$$ This definition has a very important consequence.

Multiplication Rule #1

If events $A$ and $B$ are independent, then $$P(A \cap B)=P(A) \cdot P(B)$$

Examples

Roll two fair dice. What is the probability both come up sixes?

$$P(S_1 \cap S_2)=P(S_1)\cdot P(S_2)=\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$$

Roll two fair dice. What is the probability of rolling no sixes?

$$P(N_1 \cap N_2)=P(S_1)\cdot P(S_2)=\frac{5}{6}\cdot\frac{5}{6}=\frac{25}{36}$$

Roll two fair dice. What is the probability of rolling at least $1$ six?

Let $A$ be the event "roll at least $1$ six." Then $$P(A)=1-P(N_1\cap N_2)=1-\frac{25}{36}=\frac{11}{36}$$

Verifying our Dice Probabilities

Example

Draw two cards from a well-shuffled, $52$-card deck with replacement. What is the probability of drawing two consecutive aces?

$$P(A_1 \cap A_2)=P(A_1)\cdot P(A_2)=\frac{4}{52}\cdot\frac{4}{52}=\frac{1}{13}\cdot\frac{1}{13}=\frac{1}{169}$$

Multiplication Rule #2

For any two events $A$ and $B,$ $$P(A \cap B)=P(A) \cdot P(B|A)$$

Note: Multiplication Rule #2 is just a restatement of our definition of conditional probability: $$P(B|A)=\frac{P(A \cap B)}{P(A)}$$

Another Example

Draw two cards from a well-shuffled, $52$-card deck WITHOUT replacement. What is the probability of drawing two consecutive aces?

Note: We could equivalently ask: "After shuffling, what is the probability the top two card are aces?"

$$P(A_1 \cap A_2)=P(A_1)\cdot P(A_2|A_1)=\frac{4}{52}\cdot\frac{3}{51}=\frac{1}{13}\cdot\frac{1}{17}=\frac{1}{221}$$

Yet Another Example: Billy Bob plays soccer at Southwestern Oregon Community College. He makes a goal $56\%$ of the time he shoots.

Billy Bob is going to attempt two goals in a row in the next game.
$A =$ the event Billy Bob is successful on his first attempt. $P(A)= 0.56.$
$B$ = the event Billy Bob is successful on his second attempt. $P(B)= 0.56.$
Billy Bob tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is $0.71.$

(a) What is the probability that Billy Bob makes the first goal AND the second goal?

(b) What is the probability that Billy Bob makes the first goal OR the second goal?

(a) $$P(A\cap B)=P(A)\cdot P(B|A)=0.56\cdot0.71=0.3976$$ (b) $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.56+0.56-0.3976=0.7224$$

Repeated Independent Events

What is the probability of the coin coming up "heads" on $3$ flips of a fair coin?

$$P(H_1\cap H_2 \cap H_3)=P(H_1)\cdot P(H_2)\cdot P(H_3)=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\left(\frac{1}{2}\right)^3=\frac{1}{8}$$

What is the probability of the coin coming up "heads" on $4$ flips of a fair coin?

$$P(H_1\cap H_2 \cap H_3 \cap H_4)=P(H_1)\cdot P(H_2)\cdot P(H_3)\cdot P(H_4)=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\left(\frac{1}{2}\right)^4=\frac{1}{16}$$

What is the probability of the coin coming up "heads" on $5$ flips of a fair coin?

$$P(H_1\cap H_2 \cap H_3 \cap H_4\cap H_5)=P(H_1)\cdot P(H_2)\cdot P(H_3)\cdot P(H_4)\cdot P(H_5)=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\left(\frac{1}{2}\right)^5=\frac{1}{32}$$

Multiplication Rule #1 for any number of independent events.

If events $A_1, A_2, \ldots, A_n$ are all independent events, then $$P(A_1 \cap A_2 \cap \cdots \cap A_n)=P(A_1)\cdot P(A_2) \cdots P(A_n)$$

Example

During World War II, the British found that the probability that a bomber is lost through enemy action on a mission over occupied Europe was $0.05.$

The probability that the bomber returns safely from a mission was therefore $0.95.$

It is reasonable to assume that missions are independent.

Suppose a tour of duty consists of $10$ missions. What was the probability of surviving all $10$ missions?

$$\begin{array}{ll} &P(S_1\cap S_2\cap S_3\cap S_4\cap S_5\cap S_6\cap S_7\cap S_8\cap S_9\cap S_{10})\\ &=P(S_1)\cdot P(S_2)\cdot P(S_3)\cdot P(S_4)\cdot P(S_5)\cdot P(S_6)\cdot P(S_7)\cdot P(S_8)\cdot P(S_9)\cdot P(S_{10})\\ &=(0.95)^{10}\\ &\approx 0.5987369392\\ \end{array}$$

Example

Suppose you shake up a can with $4$ six-sided dice in it and then dump them on the table.

What is the probability that at least one of the $4$ six-sided dice will be showing a six after they settle?

Let $N$ be the event "no sixes." Then $$P(N)=P(N_1\cap N_2\cap N_3\cap N_4)=P(N_1)\cdot P(N_2)\cdot P(N_3)\cdot P(N_4)=\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}=\left(\frac{5}{6}\right)^4$$ Letting the even $A$ be "at least $1$ six," we have $$P(A)=1-P(N)=1-\left(\frac{5}{6}\right)^4\approx 0.5177469136$$