Independent Events 1
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Question 1 of 5
1. Question
A multi-stage event includes three stages: spinning a spinner, tossing a coin, and rolling a dice. Find the probability of getting Yellow, Heads and the side `4`
Write fractions in the format “a/b”- (1/36)
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$Find the probability of the arrow landing on Yellow when spinning the spinner
favourable outcomes`=``1` (`1` Yellow section)total outcomes`=``3` (`3` total sections)$$ \mathsf{P(Yellow)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ Substitute values Find the probability of tossing a coin and getting Headsfavourable outcomes`=``1` (a coin has `1` Heads side)total outcomes`=``2` (a coin has `2` sides)$$ \mathsf{P(Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ Substitute values Find the probability of rolling a dice and getting `4`favourable outcomes`=``1` (a dice has `1` side with `4`)total outcomes`=``6` (a dice has `6` sides)$$\mathsf{P(4)}$$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{6}}$$ Substitute values Now, substitute the probabilities into the product rule$$\mathsf{P(Yellow)}$$ `=` $$\frac{1}{3}$$ $$\mathsf{P(Heads)}$$ `=` $$\frac{1}{2}$$ $$\mathsf{P(4)}$$ `=` $$\frac{1}{6}$$ $$\mathsf{P(Yellow\:and\:Head\:and\:4)}$$ `=` $$\mathsf{P(Yellow)}\times\mathsf{P(Heads)}\times\mathsf{P(4)}$$ Product Rule `=` $$\frac{1}{3}\times\frac{1}{2}\times\frac{1}{6}$$ Substitute values `=` $$\frac{1}{36}$$ `1/36` -
Question 2 of 5
2. Question
Two spinners are spun one after the other. Find the probability of getting the following outcomes from the `1`st and `2`nd spins respectively:
`(a)` Yellow and Green`(b)` Blue and OrangeWrite fractions in the format “a/b”-
`(a)` (1/18)`(b)` (1/9, 2/18)
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$`(a)` Find the probability of the arrow landing on Yellow and Green.Find the probability of the arrow landing on Yellow
favourable outcomes`=``1` (`1` Yellow section)total outcomes`=``3` (`3` total sections)$$ \mathsf{P(Yellow)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ Substitute values Find the probability of the arrow landing on Green
favourable outcomes`=``1` (`1` Green section)total outcomes`=``6` (`6` total sections)$$ \mathsf{P(Green)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{6}}$$ Substitute values Now, substitute the probabilities into the product rule$$\mathsf{P(Yellow)}$$ `=` $$\frac{1}{3}$$ $$\mathsf{P(Green)}$$ `=` $$\frac{1}{6}$$ $$\mathsf{P(Yellow\:and\:Green)}$$ `=` $$\mathsf{P(Yellow)}\times\mathsf{P(Green)}$$ Product Rule `=` $$\frac{1}{3}\times\frac{1}{6}$$ Substitute values `=` $$\frac{1}{18}$$ Therefore, the probability of the arrow landing on Yellow and Green is `1/18`.`(b)` Find the probability of the arrow landing on Blue and Orange.Find the probability of the arrow landing on Bluefavourable outcomes`=``1` (`1` Blue section)total outcomes`=``3` (`3` total sections)$$ \mathsf{P(Blue)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ Substitute values Find the probability of the arrow landing on Orangefavourable outcomes`=``2` (`2` Orange sections)total outcomes`=``6` (`6` total sections)$$ \mathsf{P(any\:color)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{6}}$$ Substitute values `=` $$\frac{1}{3}$$ Now, substitute the probabilities into the product rule$$\mathsf{P(Blue)}$$ `=` $$\frac{1}{3}$$ $$\mathsf{P(Orange)}$$ `=` $$\frac{1}{3}$$ $$\mathsf{P(Blue\:and\:Orange)}$$ `=` $$\mathsf{P(Blue)}\times\mathsf{P(Orange)}$$ Product Rule `=` $$\frac{1}{3}\times\frac{1}{3}$$ Substitute values `=` $$\frac{1}{9}$$ Therefore, the probability of the arrow landing on Blue and Orange is `1/9`.`(a) 1/18``(b) 1/9` -
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Question 3 of 5
3. Question
A coin is weighted in such a way that Tails would show up twice the chance of Heads. Find the probability of tossing this coin thrice and getting:`(a) 3` Tails`(b)` Tails, Heads, TailsWrite fractions in the format “a/b”-
`(a)` (8/27)`(b)` (4/27)
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$`(a)` Find the probability of getting `3` Tails.Find the probability of getting Tails. Remember that Tails has twice the chance as Headsfavourable outcomes`=``2` (T,T)total outcomes`=``3` (H,T,T)$$ \mathsf{P(Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{3}}$$ Substitute values Now, substitute this probability into the product rule$$\mathsf{P(3\:Tails)}$$ `=` $$\mathsf{P(Tails)}\times\mathsf{P(Tails)}\times\mathsf{P(Tails)}$$ Product Rule `=` $$\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}$$ Substitute values `=` $$\frac{8}{27}$$ Therefore, the probability of getting `3` Tails is `8/27`.`(b)` Find the probability of getting Tails, Heads, Tails.From part `(a)`, we have solved for the probability of getting Tails$$ \mathsf{P(Tails)} $$ `=` $$\frac{2}{3}$$ Find the probability of getting Heads. Remember that Tails has twice the chance as Headsfavourable outcomes`=``1` (H)total outcomes`=``3` (H,T,T)$$ \mathsf{P(Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ Substitute values Now, substitute the solved probabilities into the product rule$$\mathsf{P(Tails)}$$ `=` $$\frac{2}{3}$$ $$\mathsf{P(Heads)}$$ `=` $$\frac{1}{3}$$ $$\mathsf{P(Tails,\:Heads,\:Tails)}$$ `=` $$\mathsf{P(Tails)}\times\mathsf{P(Heads)}\times\mathsf{P(Tails)}$$ Product Rule `=` $$\frac{2}{3}\times\frac{1}{3}\times\frac{2}{3}$$ Substitute values `=` $$\frac{4}{27}$$ Therefore, the probability of getting Tails, Heads, Tails is `4/27`.`(a) 8/27``(b) 4/27` -
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Question 4 of 5
4. Question
One box contains cards labelled `1` to `5` and another box contains cards labelled `A` to `E`. If you draw a card from each box, find the probability of drawing:`(a)` A `2` or `4`, then an `A``(b)` An Even Number and a VowelWrite fractions in the format “a/b”-
`(a)` (2/25)`(b)` (4/25)
Hint
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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 getting a `2` or `4` then an `A`.Find the probability of getting `2` or `4`.favourable outcomes`=``2` (2,4)total outcomes`=``5` (5 number cards)$$ \mathsf{P(2\:or\:4)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{5}}$$ Substitute values Find the probability of getting `A`.favourable outcomes`=``1` (A)total outcomes`=``5` (5 letter cards)$$ \mathsf{P(A)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{5}}$$ Substitute values Now, substitute this probability into the product rule$$\mathsf{P(3\:Tails)}$$ `=` $$\mathsf{P(2\:or\:4)}\times\mathsf{P(A)}$$ Product Rule `=` $$\frac{2}{5}\times\frac{1}{5}$$ Substitute values `=` $$\frac{2}{25}$$ Therefore, the probability of getting a `2` or a `4` then an `A` is `2/25`.`(b)` Find the probability of getting an Even Number and a Vowel.Find the probability of getting an Even Number, `2` or `4`.favourable outcomes`=``2` (2,4)total outcomes`=``5` (5 number cards)$$ \mathsf{P(Even\:Number)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{5}}$$ Substitute values Find the probability of getting a Vowel, `A` or `E`.favourable outcomes`=``2` (A,E)total outcomes`=``5` (5 letter cards)$$ \mathsf{P(Vowel)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{5}}$$ Substitute values Now, substitute this probability into the product rule$$\mathsf{P(3\:Tails)}$$ `=` $$\mathsf{P(Even\:Number)}\times\mathsf{P(Vowel)}$$ Product Rule `=` $$\frac{2}{5}\times\frac{2}{5}$$ Substitute values `=` $$\frac{4}{25}$$ Therefore, the probability of getting an Even Number and a Vowel is `4/25`.`(a) 2/25``(b) 4/25` -
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Question 5 of 5
5. Question
Find the probability of rolling two dice and getting double numbers and a sum of at least `9`Write fractions in the format “a/b”- (5/108, 10/216)
Hint
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Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Product Rule
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$First, set up a lattice showing all possible sums for the two dice
This means there are `36` possible outcomesNext, find the probability of getting a double numberfavourable outcomes`=``6` (a dice has `6` numbers hence `6` possible double numbers)total outcomes`=``36`$$ \mathsf{P(double)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{6}}{\color{#007DDC}{36}}$$ Substitute values `=` $$\frac{1}{6}$$ Find the probability of getting a sum of at least `9`
favourable outcomes`=``10`(`10` dots on lattice above)total outcomes`=``36`$$ \mathsf{P(sum}≤9) $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{10}}{\color{#007DDC}{36}}$$ Substitute values `=` $$\frac{5}{18}$$ Finally, multiply the two probabilities$$\mathsf{P(double)}$$ `=` $$\frac{1}{6}$$ $$\mathsf{P(sum}≤9)$$ `=` $$\frac{5}{8}$$ $$\mathsf{P(double\:and\:sum}≤9)$$ `=` $$\mathsf{P(double)}\times\mathsf{P(sum}≤9)$$ Product Rule `=` $$\frac{1}{6}\times\frac{5}{18}$$ Substitute values `=` $$\frac{5}{108}$$ `5/108`
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)