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) Ex Probability of Independent Events 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.

Probability of Independent Events P(A and B) = P(A) · P(B) Ex Probability of Independent Events 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)

Probability of Independent Events P(A and B) = P(A) · P(B) Ex Probability of Independent Events 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

Probability of Independent Events P(A and B) = P(A) · P(B) Ex Probability of Independent Events 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

Probability of Independent Events P(A and B) = P(A) · P(B) Ex Probability of Independent Events 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

Mutually Exclusive Events P(A or B) = P(A) + P(B)

Mutually Exclusive Events P(A or B) = P(A) + P(B) Ex Mutually Exclusive Events P(A or B) = P(A) + P(B) Ex. 3 Find the probability of rolling a 3 or a 5 when rolling a die.

Mutually Exclusive Events P(A or B) = P(A) + P(B) Ex Mutually Exclusive Events 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)

Mutually Exclusive Events P(A or B) = P(A) + P(B) Ex Mutually Exclusive Events 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

Mutually Exclusive Events P(A or B) = P(A) + P(B) Ex Mutually Exclusive Events 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

Mutually Exclusive Events P(A or B) = P(A) + P(B) Ex Mutually Exclusive Events 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

Mutually Exclusive Events P(A or B) = P(A) + P(B) Ex Mutually Exclusive Events 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

Non-Mutually Exclusive Events P(A or B) = P(A) + P(B) – P(A and B)

Non-Mutually Exclusive Events P(A or B) = P(A) + P(B) – P(A and B) Ex Non-Mutually Exclusive Events 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?

Non-Mutually Exclusive Events P(A or B) = P(A) + P(B) – P(A and B) Ex Non-Mutually Exclusive Events 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)

Non-Mutually Exclusive Events P(A or B) = P(A) + P(B) – P(A and B) Ex Non-Mutually Exclusive Events 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

Non-Mutually Exclusive Events P(A or B) = P(A) + P(B) – P(A and B) Ex Non-Mutually Exclusive Events 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

Non-Mutually Exclusive Events P(A or B) = P(A) + P(B) – P(A and B) Ex Non-Mutually Exclusive Events 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