Independent Events 2
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Question 1 of 4
1. 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)` Orange and Red`(b)` Blue and any colorWrite fractions in the format “a/b”-
`(a)` (1/12)`(b)` (⅓, 1/3, 4/12)
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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 Orange and Red.Find the probability of the arrow landing on Orange<
favourable outcomes`=``1` (`1` Orange section)total outcomes`=``3` (`3` total sections)$$ \mathsf{P(Orange)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3}}$$ Substitute values Find the probability of the arrow landing on Red
favourable outcomes`=``1` (`1` Red section)total outcomes`=``4` (`4` total sections)$$ \mathsf{P(Red)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$ Substitute values Now, substitute the probabilities into the product rule$$\mathsf{P(Orange)}$$ `=` $$\frac{1}{3}$$ $$\mathsf{P(Red)}$$ `=` $$\frac{1}{4}$$ $$\mathsf{P(Orange\:and\:Red)}$$ `=` $$\mathsf{P(Orange)}\times\mathsf{P(Red)}$$ Product Rule `=` $$\frac{1}{3}\times\frac{1}{4}$$ Substitute values `=` $$\frac{1}{12}$$ Therefore, the probability of the arrow landing on Orange and Red is `1/12`.`(b)` Find the probability of the arrow landing on Blue and any color.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 any colorfavourable outcomes`=``4` (any color or section)total outcomes`=``4` (`4` total sections)$$ \mathsf{P(any\:color)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{4}}$$ Substitute values `=` $$1$$ Now, substitute the probabilities into the product rule$$\mathsf{P(Blue)}$$ `=` $$\frac{1}{3}$$ $$\mathsf{P(any\:color)}$$ `=` $$1$$ $$\mathsf{P(Blue\:and\:any\:color)}$$ `=` $$\mathsf{P(Blue)}\times\mathsf{P(any\:color)}$$ Product Rule `=` $$\frac{1}{3}\times1$$ Substitute values `=` $$\frac{1}{3}$$ Therefore, the probability of the arrow landing on Blue and any color is `1/3`.`(a) 1/12``(b) 1/3` -
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Question 2 of 4
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)` Orange and Black`(b)` Pink and redWrite fractions in the format “a/b”-
`(a)` (1/30)`(b)` (1/10, 3/30)
Hint
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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 Orange and Black.Find the probability of the arrow landing on Orange
favourable outcomes`=``1` (`1` Orange section)total outcomes`=``5` (`5` total sections)$$ \mathsf{P(Orange)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{5}}$$ Substitute values Find the probability of the arrow landing on Black
favourable outcomes`=``1` (`3` Black section)total outcomes`=``6` (`6` total sections)$$ \mathsf{P(Black)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{6}}$$ Substitute values Now, substitute the probabilities into the product rule$$\mathsf{P(Orange)}$$ `=` $$\frac{1}{5}$$ $$\mathsf{P(Black)}$$ `=` $$\frac{1}{6}$$ $$\mathsf{P(Orange\:and\:Black)}$$ `=` $$\mathsf{P(Orange)}\times\mathsf{P(Black)}$$ Product Rule `=` $$\frac{1}{5}\times\frac{1}{6}$$ Substitute values `=` $$\frac{1}{30}$$ Therefore, the probability of the arrow landing on Orange and Red is `1/30`.`(b)` Find the probability of the arrow landing on Pink and Red.Find the probability of the arrow landing on Pinkfavourable outcomes`=``1` (`1` Pink section)total outcomes`=``5` (`5` total sections)$$ \mathsf{P(Pink)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{5}}$$ Substitute values Find the probability of the arrow landing on Redfavourable outcomes`=``3` (`3` Red sections)total outcomes`=``6` (`6` total sections)$$ \mathsf{P(Red)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{6}}$$ Substitute values `=` $$\frac{1}{2}$$ Now, substitute the probabilities into the product rule$$\mathsf{P(Pink)}$$ `=` $$\frac{1}{5}$$ $$\mathsf{P(Red)}$$ `=` $$\frac{1}{2}$$ $$\mathsf{P(Pink\:and\:Red)}$$ `=` $$\mathsf{P(Pink)}\times\mathsf{P(Red)}$$ Product Rule `=` $$\frac{1}{5}\times\frac{1}{2}$$ Substitute values `=` $$\frac{1}{10}$$ Therefore, the probability of the arrow landing on Pink and Red is `1/10`.`(a) 1/30``(b) 1/10` -
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Question 3 of 4
3. Question
`4` coins were tossed at the same time. Find the probability that the `2nd` and `3rd` coins will be tails.Write fractions in the format “a/b”- (¼, 1/4, 4/16)
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
$$\mathsf{P(A\:and\:B)}=\mathsf{P(A)}\times\mathsf{P(B)}$$First, find the probability of getting Tails and getting Heads.favourable outcomes`=``1` (T)total outcomes`=``2` (H,T)$$ \mathsf{P(Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ Substitute values favourable outcomes`=``1` (H)total outcomes`=``2` (H,T)$$ \mathsf{P(Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{2}}$$ Substitute values There are four probabilities where the second and third coin would come up as Tails.`\text(H)``\text(TT)``\text(H)` `\text(H)``\text(TT)``\text(T)` `\text(T)``\text(TT)``\text(H)` `\text(T)``\text(TT)``\text(T)` Substitute the probability of getting Heads or Tails to each outcome.$$\mathsf{P(HTTH)}$$ `=` $$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$$ `=` $$\frac{1}{16}$$ $$\mathsf{P(HTTT)}$$ `=` $$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$$ `=` $$\frac{1}{16}$$ $$\mathsf{P(TTTH)}$$ `=` $$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$$ `=` $$\frac{1}{16}$$ $$\mathsf{P(TTTT)}$$ `=` $$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$$ `=` $$\frac{1}{16}$$ Finally, use the addition rule to get the total probability.$$\mathsf{P(A\:or\:B)}$$ `=` $$\mathsf{P(A)}+\mathsf{P(B)}$$ Addition Rule `=` $$\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}$$ Substitute values `=` $$\frac{4}{16}$$ `=` $$\frac{1}{4}$$ Simplify `1/4` -
Question 4 of 4
4. Question
Find the probability of rolling two dice and getting a sum of `4` and a sum greater than `10`Write fractions in the format “a/b”- (1/144, 3/432)
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 sum of `4`
favourable outcomes`=``3` (`3` dots on lattice above)total outcomes`=``36`$$ \mathsf{P(4)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{36}}$$ Substitute values `=` $$\frac{1}{12}$$ Find the probability of getting a sum greater than `10`
favourable outcomes`=``3` (`3` dots on lattice above)total outcomes`=``36`$$ \mathsf{P(sum}>10) $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{36}}$$ Substitute values `=` $$\frac{1}{12}$$ Finally, multiply the two probabilities$$\mathsf{P(4)}$$ `=` $$\frac{1}{12}$$ $$\mathsf{P(sum}>10)$$ `=` $$\frac{1}{12}$$ $$\mathsf{P(4\:and\:sum}>10)$$ `=` $$\mathsf{P(4)}\times\mathsf{P(sum}>10)$$ Product Rule `=` $$\frac{1}{12}\times\frac{1}{12}$$ Substitute values `=` $$\frac{1}{144}$$ `1/144`
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) 1
- Probability Tree (Independent) 2
- Probability Tree (Dependent)