Task Sets: Apply the general Multiplication Rule - Common Core Math - HS Statistics and Probability

Warm-Up: Product Rule and Conditional Probability

These problems review the prior ideas the new rule builds on.

For independent events, the product rule from earlier is $P(A \text{ and } B) = P(A) \cdot P(B)$ . Two cards are drawn from a 52-card deck without replacement. Why might this simple product rule give the wrong joint probability here?

Because the draws are dependent: removing the first card changes the probabilities for the second draw, so the second factor is no longer the original $P(B)$ .

Because probabilities can never be multiplied when cards are involved.

Because the deck has too many cards for the rule to apply.

Because $P(A) \cdot P(B)$ always overcounts by exactly one card.

The conditional-probability formula is $P(B \mid A) = \dfrac{P(A \text{ and } B)}{P(A)}$ . Suppose $P(A) = 0.5$ and $P(A \text{ and } B) = 0.2$ . Find $P(B \mid A)$ as a decimal.

The general Multiplication Rule has two symmetric forms: $P(A) \cdot P(B \mid A)$ and $P(B) \cdot P(A \mid B)$ . What is true about these two expressions?

They always give the same value, the joint probability $P(A \text{ and } B)$ .

They give different answers, so you must memorize which to use.

They are equal only when $A$ and $B$ are independent.

The first is the joint probability and the second is the conditional probability.

Fluency: Apply the Rule

Apply the general Multiplication Rule. For without-replacement draws, update both the favorable count and the total.

Given $P(A) = 0.4$ and $P(B \mid A) = 0.5$ , use the general Multiplication Rule to find $P(A \text{ and } B)$ . Give a decimal.

Two cards are drawn from a standard 52-card deck without replacement. Find $P(\text{both kings})$ by filling each blank. First king: $P = \dfrac{4}{52}$ . After a king is removed, ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ kings remain among ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ cards, so the second factor is $\dfrac{3}{\text{\hspace{0.2em}\fbox{\phantom{000000}}\hspace{0.2em}}}$ . The joint probability, in lowest terms, is $\dfrac{1}{\text{\hspace{0.2em}\fbox{\phantom{000000}}\hspace{0.2em}}}$ .

kings remaining:

cards remaining:

second-draw denominator:

denominator of P(both kings) in lowest terms:

A bag has 3 red and 2 blue marbles. Two are drawn without replacement. Find $P(\text{red, then blue})$ , the probability the first is red and the second is blue. Give your answer as a fraction in lowest terms.

On a tree diagram for the marble bag (3 red, 2 blue, drawn without replacement), the red-then-blue path has branch probabilities $\dfrac{3}{5}$ and $\dfrac{2}{4}$ . How do you combine them to get the probability of that single path?

Multiply along the path: $\dfrac{3}{5} \cdot \dfrac{2}{4} = \dfrac{3}{10}$ .

Add along the path: $\dfrac{3}{5} + \dfrac{2}{4} = \dfrac{11}{10}$ .

Subtract: $\dfrac{3}{5} - \dfrac{2}{4} = \dfrac{1}{10}$ .

Take the larger branch probability, $\dfrac{3}{5}$ .

A jar holds 4 green and 6 yellow candies. Two are eaten one after another (without replacement). Find $P(\text{both green})$ as a fraction in lowest terms.

Exercises: Apply the General Multiplication Rule

Warm-Up: Product Rule and Conditional Probability

Fluency: Apply the Rule