Independence and the multiplication rule

About this lesson

Geometry › Conditional probability

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Independence and the multiplication rule

Warm-up

Three quick questions from earlier lessons.

Discover

A card is drawn from a deck. Does knowing it is a heart change the probability that it is a king?

  1. No
  2. Yes

Key idea

Independent events do not influence each other.

AA and BB are independent if P(A∣B)=P(A)P(A \mid B) = P(A), equivalently P(A and B)=P(A) P(B)P(A \text{ and } B) = P(A)\,P(B). In general P(A and B)=P(A) P(B∣A)P(A \text{ and } B) = P(A)\,P(B \mid A). Draws without replacement are dependent.

Worked example

Two cards drawn without replacement: P(both aces).

P(ace)=452P(\text{ace}) = \tfrac{4}{52}
First draw.
P(ace∣ace)=351P(\text{ace} \mid \text{ace}) = \tfrac{3}{51}
3 aces left of 51.
452⋅351=1221\tfrac{4}{52} \cdot \tfrac{3}{51} = \tfrac{1}{221}
Multiply.

Your turn, guided

Your turn: two independent events with probabilities 0.5 and 0.3. P(both).

0.150.15
0.5 × 0.3.

Practice ★

  1. Two independent machines fail with probabilities 0.8 and 0.7. Find the probability that both fail.
  2. Two independent machines fail with probabilities 0.7 and 0.4. Find the probability that both fail.
  1. P(A) = 3/10, P(B) = 6/10 and P(A and B) = 18/100. Are A and B independent?
    1. Yes: P(A and B) = P(A)·P(B)
    2. No, because P(A and B) > 0
    3. It cannot be decided

Watch out!

Practice ★★

  1. A bag has 5 red and 3 blue marbles. Two are drawn without replacement. Find P(both red).
  2. A bag has 6 red and 7 blue marbles. Two are drawn without replacement. Find P(both red).

Exit ticket

Three questions to check you have got it.

  1. Two independent machines fail with probabilities 0.8 and 0.7. Find the probability that both fail.
  2. Two independent machines fail with probabilities 0.7 and 0.4. Find the probability that both fail.
  3. Two independent machines fail with probabilities 0.3 and 0.5. Find the probability that both fail.

Summary card

Independence

  • Independent: P(A and B)=P(A)P(B)P(A \text{ and } B) = P(A)P(B).
  • In general: P(A)P(B∣A)P(A)P(B \mid A).

Watch out: without replacement, update the second probability.

End of the lesson

Next lesson: Tree diagrams and reverse conditioning Back to the chapter

Answers: Independence and the multiplication rule

A

  1. 1425\frac{14}{25}
  2. 725\frac{7}{25}

B

  1. Yes: P(A and B) = P(A)·P(B)

C

  1. 514\frac{5}{14}
  2. 526\frac{5}{26}

D

  1. 1425\frac{14}{25}
  2. 725\frac{7}{25}
  3. 320\frac{3}{20}

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