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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 ) - PowerPoint PPT Presentation
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12.5 – Probability of Compound Events
Probability of Independent EventsP(A and B) = P(A) · P(B)
Probability of Independent EventsP(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 EventsP(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 EventsP(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 EventsP(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 EventsP(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 EventsP(A and B) = P(A) · P(B following A)
Probability of Dependent EventsP(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 EventsP(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 EventsP(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 EventsP(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 EventsP(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 EventsP(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 EventsP(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 EventsP(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 EventsP(A or B) = P(A) + P(B)
Mutually Exclusive EventsP(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 EventsP(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 EventsP(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 EventsP(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 EventsP(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 EventsP(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 EventsP(A or B) = P(A) + P(B) – P(A and B)
Non-Mutually Exclusive EventsP(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 EventsP(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 EventsP(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 EventsP(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 EventsP(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
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