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Probability Tree (Independent Events) 2Probability Tree (Independent Events) 2
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Question 1 of 5
1. Question
A couple plan to have three kids. Find the probability of having:`(a) 2` Girls and `1` Boy (in any order)`(b) 3` of the same genderWrite fractions in the format “a/b”-
`(a)` (⅜, 3/8)`(b)` (¼, 1/4, 2/8)
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$`(a)` Find the probability of having `2` Girls and `1` Boy.First, set up a probability tree showing all possible genders of `3` kids
Find the probability of all outcomes$$ \mathsf{P(Girl)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` out of `2` possible genders $$ \mathsf{P(Boy)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` out of `2` possible genders Mark each outcome with its corresponding probability
Pick the branches that lead to `2` Girls and `1` Boy
First branch (BGG):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Second branch (GBG):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Third branch (GGB):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Finally, add the solved probability for each branch`1/8``+``1/8``+``1/8` `=` `3/8` Addition Rule Therefore, the probability of having `2` Girls and `1` Boy is `3/8``(b)` Find the probability of having `3` of the same gender.Use the probability tree from part `(a)` and pick the branches that lead to `3` of the same gender then multiply the probabilities along it
First branch (BBB):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Second branch (GGG):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Finally, add the solved probability for each branch`1/8``+``1/8` `=` `2/8` `=` `1/4` Addition Rule Therefore, the probability of having `3` of the same gender is `1/4``(a) 3/8``(b) 1/4` -
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Question 2 of 5
2. Question
A box contains `5` balls, `3` of which are Yellow and `2` are Blue. Find the probability of drawing `3` balls at random and getting:`(a) 3` Yellow`(b)` at least `1` BlueWrite fractions in the format “a/b”-
`(a)` (27/125)`(b)` (98/125)
Hint
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Complementary
Probability$$\mathsf{P(\dot{E})}=1-\mathsf{P(E)}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$`(a)` Find the probability of drawing `3` Yellow balls.First, set up a probability tree showing all possible outcomes of drawing `3` balls
Find the probability of all outcomes$$ \mathsf{P(Yellow)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{5}}$$ `3` Yellow balls out of `5` balls $$ \mathsf{P(Blue)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{5}}$$ `2` Blue balls out of `5` balls Mark each outcome with its corresponding probability
Finally, pick the branch that lead to `3` Yellow and multiply the probabilities along it
`3/5``times``3/5``times``3/5` `=` `27/125` Product Rule Therefore, the probability of drawing at `3` Yellow balls is `27/125``(b)` Find the probability of drawing at least `1` Blue ball.Notice that all outcomes have at least `1` Blue except YYYHence, we can simply get the complement of P(YYY)
$$ \mathsf{P(\dot{YYY})} $$ `=` $$1-\mathsf{P(YYY)}$$ Complementary Probability `=` `1-27/125` Substitute value `=` `125/125-27/125` `=` `98/125` Therefore, the probability of drawing at least `1` Blue is `98/125``(a) 27/125``(b) 98/125` -
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Question 3 of 5
3. Question
A box contains `5` balls, `3` of which are Yellow and `2` are Blue. Find the probability of drawing `3` balls at random and getting:Exactly `2` BlueWrite fractions in the format “a/b”- (36/125)
Hint
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Complementary
Probability$$\mathsf{P(\dot{E})}=1-\mathsf{P(E)}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$`(a)` Find the probability of drawing exactly `2` Blue balls.First, set up a probability tree showing all possible outcomes of drawing `3` balls
Find the probability of all outcomes$$ \mathsf{P(Yellow)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{5}}$$ `3` Yellow balls out of `5` balls $$ \mathsf{P(Blue)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{5}}$$ `2` Blue balls out of `5` balls Mark each outcome with its corresponding probability
Finally, pick the branches that lead to exactly `2` Blue and multiply the probabilities along it
First branch (YBB):`3/5``times``2/5``times``2/5` `=` `12/125` Product Rule Second branch (BYB):`2/5``times``3/5``times``2/5` `=` `12/125` Product Rule Third branch (BBY):`2/5``times``2/5``times``3/5` `=` `12/125` Product Rule Finally, add the solved probability for each branch`12/125``+``12/125``+``12/125` `=` `36/125` Addition Rule Therefore, the probability of drawing exactly `2` blue balls is `36/125``36/125` -
Question 4 of 5
4. Question
A person driving on the road goes through 3 sets of traffic lights. The probability of driving through a green light is `0.6`.What is the probability that the driver gets:Exactly `2` Green LightsAnswer to `3` decimal places- (0.432)
Hint
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Complementary
Probability$$\mathsf{P(\dot{E})}=1-\mathsf{P(E)}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$First, set up a probability tree showing all possible outcomes of getting exactly `2` green lights
We are given the probability of going through a Green Light from the question `(0.6)`. Therefore the probability of going through a Red Light would be `0.4`.`P(\text(Green Light))` `=` `0.6` `P(\text(Red))` `=` `0.4` `1-0.6=0.4` Mark each outcome with its corresponding probability
Finally, pick the branches that lead to exactly `2` Green Lights and multiply the probabilities along it
First branch (GGR):`0.6``times``0.6``times``0.4` `=` `0.144` Product Rule Second branch (GRG):`0.6``times``0.4``times``0.6` `=` `0.144` Product Rule Third branch (RGG):`0.4``times``0.6``times``0.4` `=` `0.144` Product Rule Finally, add the solved probability for each branch`0.144``+``0.144``+``0.144` `=` `0.432` Addition Rule Therefore, the probability of getting exactly `2` Green Lights is `0.432``0.432` -
Question 5 of 5
5. Question
An athlete competes in a triathlon that involves swimming, cycling and running.His chances of winning each event are:Swimming: `2/3`Cycling: `1/4`Running: `1/2`What is the probability of this athlete only losing one event?Write fractions in the format “a/b”- (⅜, 3/8, 9/24)
Hint
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Correct!
Incorrect
Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Complementary
Probability$$\mathsf{P(\dot{E})}=1-\mathsf{P(E)}$$Addition Rule
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$First, set up a probability tree showing all possible outcomes of losing `1` event
We are given the probabilities of winning each event from the question`P(\text(Win the Swimming Event))` `=` `2/3` `P(\text(Win the Cycling Event))` `=` `1/4` `P(\text(Win the Running Event))` `=` `1/2` Mark each outcome with its corresponding probability
Finally, pick the branches that lead to outcomes where the athlete loses `1` event and multiply the probabilities along it
First branch (WWL):`2/3``xx``1/4``xx``1/2` `=` `2/24` Product Rule Second branch (WLW):`2/3``xx``3/4``xx``1/2` `=` `6/24` Product Rule Third branch (LWW):`1/3``xx``1/4``xx``1/2` `=` `1/24` Product Rule Finally, add the solved probability for each branch`2/24``+``6/24``+``1/24` `=` `9/24` Addition Rule `=` `3/8` Simplify Therefore, the probability of the athlete losing `1` event is `3/8``3/8`
Quizzes
- Simple Probability (Theoretical) 1
- Simple Probability (Theoretical) 2
- Simple Probability (Theoretical) 3
- Simple Probability (Theoretical) 4
- Complementary Probability
- Compound Events (Addition Rule) 1
- Compound Events (Addition Rule) 2
- Venn Diagrams (Mutually Inclusive)
- Independent Events 1
- Independent Events 2
- Dependent Events (Conditional Probability)
- Probability Tree (Independent Events) 1
- Probability Tree (Independent Events) 2
- Probability Tree (Dependent Events)