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Probability Tree (Independent)Probability Tree (Independent)
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Question 1 of 6
1. Question
Find the probability of tossing a normal coin twice and getting:`(a) 2` Tails`(b)` Heads and Tails (in any order)Write fractions in the format “a/b”-
`(a)` (1/4)`(b)` (1/2, 2/4)
Hint
Help VideoCorrect
Well Done!
Incorrect
Help VideoProbability 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 getting `2` Tails.First, set up a probability tree showing all possible outcomes of tossing a coin twice
Find the probability of all outcomes$$ \mathsf{P(Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Heads out of `2` sides $$ \mathsf{P(Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Tails out of `2` sides Mark each outcome with its corresponding probability
Finally, solve for the probability by picking the branch that leads to `2` Tails and multiplying the probabilities along it
`1/2``times``1/2` `=` `1/4` Product Rule Therefore, the probability of throwing Tails twice is `1/4``(b)` Find the probability of getting Heads and Tails.Use the probability tree from part `(a)` and pick the branches that lead to Heads and Tails then multiply the probabilities along it
First branch (HT):`1/2``times``1/2` `=` `1/4` Product Rule Second branch (TH):`1/2``times``1/2` `=` `1/4` Product Rule Finally, add the solved probability for each branch`1/4``+``1/4` `=` `2/4` `=` `1/2` Addition Rule Therefore, the probability of getting Heads and Tails is `1/2``(a) 1/4``(b) 1/2` -
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Question 2 of 6
2. Question
Find the probability of tossing a normal coin thrice and getting:`(a)` Heads, Tails, Tails (in that order)`(b) 2` Tails and Heads (in any order)Write fractions in the format “a/b”-
`(a)` (1/8)`(b)` (3/8)
Hint
Help VideoCorrect
Nice Job!
Incorrect
Help VideoProbability 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 getting Heads, Tails, Tails.First, set up a probability tree showing all possible outcomes of tossing a coin thrice
Find the probability of all outcomes$$ \mathsf{P(Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Heads out of `2` sides $$ \mathsf{P(Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Tails out of `2` sides Mark each outcome with its corresponding probability
Finally, solve for the probability by picking the branch that leads to Heads, Tails, Tails and multiplying the probabilities along it
`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Therefore, the probability of throwing Heads, Tails, Tails twice in that order is `1/8``(b)` Find the probability of getting `2` Tails and `1` Heads.Use the probability tree from part `(a)` and pick the branches that lead to `2` Tails and `1` Heads then multiply the probabilities along it
First branch (HTT):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Second branch (THT):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Third branch (TTH):`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 getting `2` Tails and `1` Heads is `3/8``(a) 1/8``(b) 3/8` -
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Question 3 of 6
3. Question
Find the probability of tossing a normal coin thrice and getting:`(a) 3` of a kind`(b)` at least `2` HeadsWrite fractions in the format “a/b”-
`(a)` (1/4, 2/8)`(b)` (1/2, 4/8)
Hint
Help VideoCorrect
Excellent!
Incorrect
Help VideoProbability 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 getting `3` of a kind.First, set up a probability tree showing all possible outcomes of tossing a coin thrice
Find the probability of all outcomes$$ \mathsf{P(Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Heads out of `2` sides $$ \mathsf{P(Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Tails out of `2` sides Mark each outcome with its corresponding probability
Pick the branches that lead to `3` of a kind then multiply the probabilities along it
First branch (HHH):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Second branch (TTT):`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 getting `3` of a kind is `1/4``(b)` Find the probability of getting at least `2` Heads.Use the probability tree from part `(a)` and pick the branches that lead to at least `2` Heads then multiply the probabilities along it
First branch (HHT):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Second branch (HTH):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Third branch (HHT):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Fourth branch (HHH):`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``+``1/8` `=` `4/8` `=` `1/2` Addition Rule Therefore, the probability of getting at least `2` Heads is `1/2``(a) 1/4``(b) 1/2` -
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Question 4 of 6
4. Question
Find the probability of tossing a normal coin thrice and getting `3` Heads or Heads, Tails, Heads in that order.Write fractions in the format “a/b”- (1/4, 2/8)
Hint
Help VideoCorrect
Fantastic!
Incorrect
Help VideoProbability 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
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$First, set up a probability tree showing all possible outcomes of tossing a coin thrice
Find the probability of all outcomes$$ \mathsf{P(Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Heads out of `2` sides $$ \mathsf{P(Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ `1` Tails out of `2` sides Mark each outcome with its corresponding probability
Pick the branches that lead to the preferred outcomes then multiply the probabilities along it
First branch (HHH):`1/2``times``1/2``times``1/2` `=` `1/8` Product Rule Second branch (HTH):`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 getting `3` Heads and Heads, Tails, Heads is `1/4``1/4` -
Question 5 of 6
5. 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)` (3/8)`(b)` (1/4, 2/8)
Hint
Help VideoCorrect
Keep Going!
Incorrect
Help VideoProbability 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 sex.Use the probability tree from part `(a)` and pick the branches that lead to `3` of the same sex 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 sex is `1/4``(a) 3/8``(b) 1/4` -
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Question 6 of 6
6. 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
Help VideoCorrect
Correct!
Incorrect
Help VideoProbability 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` -
Quizzes
- Simple Probability 1
- Simple Probability 2
- Simple Probability 3
- Simple Probability 4
- Complementary Probability 1
- Compound Events 1
- Compound Events 2
- Venn Diagrams (Non Mutually Exclusive)
- Independent Events 1
- Independent Events 2
- Dependent Events (Conditional Probability)
- Probability Tree (Independent)
- Probability Tree (Dependent)