3.2 Conditional Probability and the Multiplication Rule

3.2 Conditional Probability and the Multiplication Rule

Assessment

Flashcard

Mathematics

9th - 12th Grade

Hard

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14 questions

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1.

FLASHCARD QUESTION

Front

What is conditional probability?

Back

Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B), which means the probability of event A occurring given that event B has occurred.

2.

FLASHCARD QUESTION

Front

What is the multiplication rule in probability?

Back

The multiplication rule states that the probability of two independent events A and B occurring together is the product of their individual probabilities: P(A and B) = P(A) * P(B).

3.

FLASHCARD QUESTION

Front

How do you calculate the probability of getting 4 heads when tossing a coin 4 times?

Back

The probability of getting heads in one toss is 0.5. For 4 heads in 4 tosses, use the formula: P(4 heads) = (0.5)^4 = 0.0625.

4.

FLASHCARD QUESTION

Front

What is the probability of selecting a defective unit from a shipment?

Back

The probability of selecting a defective unit is calculated by dividing the number of defective units by the total number of units. For example, if there are 3 defective units in 250, P(defective) = 3/250 = 0.012.

5.

FLASHCARD QUESTION

Front

How do you find the probability of receiving no defective units when selecting multiple items?

Back

To find the probability of receiving no defective units, use the formula: P(no defective) = (number of non-defective units / total units) * (number of non-defective units - 1 / total units - 1) * ... for the number of selections.

6.

FLASHCARD QUESTION

Front

What is the formula for finding the probability of two dependent events?

Back

For two dependent events A and B, the probability is calculated as: P(A and B) = P(A) * P(B|A), where P(B|A) is the probability of B given that A has occurred.

7.

FLASHCARD QUESTION

Front

How do you calculate the probability of selecting an ace of diamonds and then a jack of diamonds without replacement?

Back

P(ace of diamonds) = 1/52. After selecting the ace, P(jack of diamonds) = 1/51. Therefore, P(ace and jack) = (1/52) * (1/51) = 0.000377.

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