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# 12.5 – Probability of Compound Events - PowerPoint PPT Presentation

12.5 – Probability of Compound Events. Probability of Independent Events P( A and B ) = P( A ) · P( B ). Probability of Independent Events P( A and B ) = P( A ) · P( B )

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### 12.5 – Probability of Compound Events

P(A and B) = P(A) · P(B)

P(A and B) = P(A) · P(B)

Ex. 1 A bag contains 6 black marbles, 9 blue marbles, 4 yellow marbles, and 2 green marbles. A marble is selected, replaced, and a second marble is selected. Find the probability of selecting a black marble, ehtn a yellow marble.

P(A and B) = P(A) · P(B)

Ex. 1 A bag contains 6 black marbles, 9 blue marbles, 4 yellow marbles, and 2 green marbles. A marble is selected, replaced, and a second marble is selected. Find the probability of selecting a black marble, ehtn a yellow marble.

P(black, yellow) = P(black) · P(yellow)

P(A and B) = P(A) · P(B)

Ex. 1 A bag contains 6 black marbles, 9 blue marbles, 4 yellow marbles, and 2 green marbles. A marble is selected, replaced, and a second marble is selected. Find the probability of selecting a black marble, ehtn a yellow marble.

P(black, yellow) = P(black) · P(yellow)

= 6_ . 4_

21 21

P(A and B) = P(A) · P(B)

Ex. 1 A bag contains 6 black marbles, 9 blue marbles, 4 yellow marbles, and 2 green marbles. A marble is selected, replaced, and a second marble is selected. Find the probability of selecting a black marble, ehtn a yellow marble.

P(black, yellow) = P(black) · P(yellow)

= 6_ . 4_ = 24_

21 21 441

P(A and B) = P(A) · P(B)

Ex. 1 A bag contains 6 black marbles, 9 blue marbles, 4 yellow marbles, and 2 green marbles. A marble is selected, replaced, and a second marble is selected. Find the probability of selecting a black marble, ehtn a yellow marble.

P(black, yellow) = P(black) · P(yellow)

= 6_ . 4_ = 24_ ≈ 5.4%

21 21 441

Probability of Dependent Events

P(A and B) = P(A) · P(B following A)

Probability of Dependent Events

P(A and B) = P(A) · P(B following A)

Ex. 2 Cynthia randomly draws three cards from a standard deck one at a time without replacement. Find the probability that the first card is a diamond, the second a spade, and the third a diamond.

Probability of Dependent Events

P(A and B) = P(A) · P(B following A)

Ex. 2 Cynthia randomly draws three cards from a standard deck one at a time without replacement. Find the probability that the first card is a diamond, the second a spade, and the third a diamond.

P(d, s, d) = P(diamond) · P(spade) · P(diamond)

Probability of Dependent Events

P(A and B) = P(A) · P(B following A)

Ex. 2 Cynthia randomly draws three cards from a standard deck one at a time without replacement. Find the probability that the first card is a diamond, the second a spade, and the third a diamond.

P(d, s, d) = P(diamond) · P(spade) · P(diamond)

= 13_

52

Probability of Dependent Events

P(A and B) = P(A) · P(B following A)

Ex. 2 Cynthia randomly draws three cards from a standard deck one at a time without replacement. Find the probability that the first card is a diamond, the second a spade, and the third a diamond.

P(d, s, d) = P(diamond) · P(spade) · P(diamond)

= 13_ . 13_

52 51

Probability of Dependent Events

P(A and B) = P(A) · P(B following A)

Ex. 2 Cynthia randomly draws three cards from a standard deck one at a time without replacement. Find the probability that the first card is a diamond, the second a spade, and the third a diamond.

P(d, s, d) = P(diamond) · P(spade) · P(diamond)

= 13_ . 13_ . 12_

52 51 50

Probability of Dependent Events

P(A and B) = P(A) · P(B following A)

Ex. 2 Cynthia randomly draws three cards from a standard deck one at a time without replacement. Find the probability that the first card is a diamond, the second a spade, and the third a diamond.

P(d, s, d) = P(diamond) · P(spade) · P(diamond)

= 13_ . 13_ . 12_

52 51 50

= 1_ . 13_ . 6_

4 51 25

Probability of Dependent Events

P(A and B) = P(A) · P(B following A)

Ex. 2 Cynthia randomly draws three cards from a standard deck one at a time without replacement. Find the probability that the first card is a diamond, the second a spade, and the third a diamond.

P(d, s, d) = P(diamond) · P(spade) · P(diamond)

= 13_ . 13_ . 12_

52 51 50

= 1_ . 13_ . 6_ = 13_

4 51 25 850

Probability of Dependent Events

P(A and B) = P(A) · P(B following A)

Ex. 2 Cynthia randomly draws three cards from a standard deck one at a time without replacement. Find the probability that the first card is a diamond, the second a spade, and the third a diamond.

P(d, s, d) = P(diamond) · P(spade) · P(diamond)

= 13_ . 13_ . 12_

52 51 50

= 1_ . 13_ . 6_ = 13_ ≈ 1.5%

4 51 25 850

P(A or B) = P(A) + P(B)

P(A or B) = P(A) + P(B)

Ex. 3 Find the probability of rolling a 3 or a 5 when rolling a die.

P(A or B) = P(A) + P(B)

Ex. 3 Find the probability of rolling a 3 or a 5 when rolling a die.

P(3 or 5) = P(3) + P(5)

P(A or B) = P(A) + P(B)

Ex. 3 Find the probability of rolling a 3 or a 5 when rolling a die.

P(3 or 5) = P(3) + P(5)

= 1_ + 1_

6 6

P(A or B) = P(A) + P(B)

Ex. 3 Find the probability of rolling a 3 or a 5 when rolling a die.

P(3 or 5) = P(3) + P(5)

= 1_ + 1_ = 2_

6 6 6

P(A or B) = P(A) + P(B)

Ex. 3 Find the probability of rolling a 3 or a 5 when rolling a die.

P(3 or 5) = P(3) + P(5)

= 1_ + 1_ = 2_ = 1

6 6 6 3

P(A or B) = P(A) + P(B)

Ex. 3 Find the probability of rolling a 3 or a 5 when rolling a die.

P(3 or 5) = P(3) + P(5)

= 1_ + 1_ = 2_ = 1 ≈ 33%

6 6 6 3

P(A or B) = P(A) + P(B) – P(A and B)

P(A or B) = P(A) + P(B) – P(A and B)

Ex. 4 Out of 5200 households surveyed, 2107 had a dog, 807 had a cat, and 303 had both. What is the probability that a randomly selected household has either a dog or a cat?

P(A or B) = P(A) + P(B) – P(A and B)

Ex. 4 Out of 5200 households surveyed, 2107 had a dog, 807 had a cat, and 303 had both. What is the probability that a randomly selected household has either a dog or a cat?

P(dog or cat) = P(dog) + P(cat) – P(both)

P(A or B) = P(A) + P(B) – P(A and B)

Ex. 4 Out of 5200 households surveyed, 2107 had a dog, 807 had a cat, and 303 had both. What is the probability that a randomly selected household has either a dog or a cat?

P(dog or cat) = P(dog) + P(cat) – P(both)

= 2107_ + 807_ – 303_

5200 5200 5200

P(A or B) = P(A) + P(B) – P(A and B)

Ex. 4 Out of 5200 households surveyed, 2107 had a dog, 807 had a cat, and 303 had both. What is the probability that a randomly selected household has either a dog or a cat?

P(dog or cat) = P(dog) + P(cat) – P(both)

= 2107_ + 807_ – 303_ = 2611

5200 5200 5200 5200

P(A or B) = P(A) + P(B) – P(A and B)

Ex. 4 Out of 5200 households surveyed, 2107 had a dog, 807 had a cat, and 303 had both. What is the probability that a randomly selected household has either a dog or a cat?

P(dog or cat) = P(dog) + P(cat) – P(both)

= 2107_ + 807_ – 303_ = 2611 ≈ 50%

5200 5200 5200 5200