Independent Events 2
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 4 questions completed
Questions:
- 1
- 2
- 3
- 4
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- Answered
- Review
-
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)
Hint
Help VideoCorrect
Well Done!
Incorrect
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` -
-
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
Help VideoCorrect
Excellent!
Incorrect
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` -
-
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
Help VideoCorrect
Keep Going!
Incorrect
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
Help VideoCorrect
Keep Going!
Incorrect
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`