Compound Events 1
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 8 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
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
- 5
- 6
- 7
- 8
- Answered
- Review
-
Question 1 of 8
1. Question
A normal six-sided dice is rolled. Identify which of the following are mutually exclusive events:Hint
Help VideoCorrect
Great Work!
Incorrect
Events are mutually exclusive if they cannot occur simultaneously.Rolling the dice and getting an Even or Odd number.Compare the results for each separate eventSample Space of rolling a dice: `1,2,3,4,5,6`Even: `2,4,6` Odd: `1,3,5` None of the results occurred simultaneously or is repeated. Hence, these are mutually exclusive events.Rolling the dice and getting an Even or `>``5` number of dots.Compare the results for each separate eventSample Space of rolling a dice: `1,2,3,4,5,6`Even: `2,4,``6` `>``5`: `6` `6` dots occurred simultaneously and is repeated on the two events. Hence, these are NOT mutually exclusive events.Getting an Even or Odd number of dots is a mutually exclusive event -
Question 2 of 8
2. Question
A jar contains `9` Red, `4` Green and `2` Black marbles. Find the probability of drawing a marble at random and getting:`(a)` Red or Green`(b)` Red or BlackWrite fractions in the format “a/b”-
`(a)` (13/15)`(b)` (11/15)
Hint
Help VideoCorrect
Keep Going!
Incorrect
Addition Rule (Mutually Exclusive Events)
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$`(a)` Find the probability drawing a Red or Green marble.Start by finding the probability of drawing a Red marblefavourable outcomes`=``9` (`9` Red)total outcomes`=``15` (`9` Red, `4` Green, `2` Black)$$ \mathsf{P(Red)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{9}}{\color{#007DDC}{15}}$$ Substitute values Next, find the probability of drawing a Green marblefavourable outcomes`=``4` (`4` Green)total outcomes`=``15` (`9` Red, `4` Green, `2` Black)$$ \mathsf{P(Green)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{15}}$$ Substitute values Finally, add the two probabilities$$ \mathsf{P(Red\:or\:Green)} $$ `=` $$\mathsf{P(Red)}+\mathsf{P(Green)}$$ Addition Rule `=` $$\frac{9}{15}+\frac{4}{15}$$ Substitute values `=` $$\frac{13}{15}$$ `(b)` Find the probability drawing a Red or Black marble.From part `(a)`, we have solved the probability of drawing a Red marble$$ \mathsf{P(Red)} $$ `=` $$\frac{9}{15}$$ Next, find the probability of drawing a Black marblefavourable outcomes`=``2` (`2` Black)total outcomes`=``15` (`9` Red, `4` Green, `2` Black)$$ \mathsf{P(Black)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{15}}$$ Substitute values Finally, add the two probabilities$$ \mathsf{P(Red\:or\:Black)} $$ `=` $$\mathsf{P(Red)}+\mathsf{P(Black)}$$ Addition Rule `=` $$\frac{9}{15}+\frac{2}{15}$$ Substitute values `=` $$\frac{11}{15}$$ `(a) 13/15``(b) 11/15` -
-
Question 3 of 8
3. Question
Find the probability of drawing from a standard deck of cards and getting:`(a)` Black `8` or Hearts`(b)` King or RedWrite fractions in the format “a/b”-
`(a)` (15/52)`(b)` (7/13, 28/52)
Hint
Help VideoCorrect
Excellent!
Incorrect
Addition Rule (Mutually Exclusive Events)
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$`(a)` Find the probability of drawing a Black `8` or Hearts.Start by finding the probability of drawing a Black `8`favourable outcomes`=``2` (`8` of Spades, `8` of Clubs)total outcomes`=``52` (a standard deck has `52` cards)$$ \mathsf{P(Black\:8)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{52}}$$ Substitute values Next, find the probability of drawing a Hearts cardfavourable outcomes`=``13` (`13` Hearts cards)total outcomes`=``52` (a standard deck has `52` cards)$$ \mathsf{P(Hearts)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{13}}{\color{#007DDC}{52}}$$ Substitute values Finally, add the two probabilities$$ \mathsf{P(Black\:8\:or\:Hearts)} $$ `=` $$\mathsf{P(Black\:8)}+\mathsf{P(Hearts)}$$ Addition Rule `=` $$\frac{2}{52}+\frac{13}{52}$$ Substitute values `=` $$\frac{15}{52}$$ `(b)` Find the probability of drawing a King or Red card.Start by finding the probability of drawing a Kingfavourable outcomes`=``4` (each of the suits have a King card)total outcomes`=``52` (a standard deck has `52` cards)$$ \mathsf{P(King)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{4}}{\color{#007DDC}{52}}$$ Substitute values Next, find the probability of drawing a Red cardfavourable outcomes`=``26` (`13` Hearts, `13` Diamonds)total outcomes`=``52` (a standard deck has `52` cards)$$ \mathsf{P(Red)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{26}}{\color{#007DDC}{52}}$$ Substitute values Remember that the events need to be mutually exclusive which means the same card can’t be counted twice.`2` of the cards are counted twice, which are King of Hearts and King of DiamondsHence, subtract `2/52` from the final probability.Finally, solve for the final probability$$ \mathsf{P(King\:or\:Red)} $$ `=` $$\mathsf{P(King)}+\mathsf{P(Red)}-\frac{2}{52}$$ Addition Rule `=` $$\frac{4}{52}+\frac{26}{52}-\frac{2}{52}$$ Substitute values `=` $$\frac{28}{52}$$ `=` $$\frac{7}{13}$$ `(a) 15/52``(b) 7/13` -
-
Question 4 of 8
4. Question
Find the probability of rolling `2` normal six-sided dice and getting:`(a)` Sum of `10``(b)` Sum of `7`Write fractions in the format “a/b”-
`(a)` (1/12, 3/36)`(b)` (1/6, 6/36)
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 getting a Sum of `10`.Set up a lattice showing all possible sums for the two dice
This means there are `36` possible outcomesNow, count how many times we can have a sum of `10`
This means there are `3` times that we can have a sum of `10`Finally, find the probability of getting a sum of `10`favourable outcomes`=``3`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 `=` $$\frac{1}{12}$$ `(b)` Find the probability of getting a Sum of `7`.Use the lattice from part `(a)` and count how many times we can have a sum of `7`
This means there are `6` times that we can have a sum of `7`Finally, find the probability of getting a sum of `7`favourable outcomes`=``6`total outcomes`=``36`$$ \mathsf{P(sum}\:7) $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{6}}{\color{#007DDC}{36}}$$ Substitute values `=` $$\frac{1}{6}$$ `(a) 1/12``(b) 1/6` -
-
Question 5 of 8
5. Question
A normal six-sided dice is rolled and a fair coin is tossed. Find the probability of getting an Even number of dots and Heads.Write fractions in the format “a/b”- (¼, 1/4, 3/12)
Hint
Help VideoCorrect
Keep Going!
Incorrect
Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Set up a lattice showing all possible results for rolling a dice and tossing a coin
This means there are `12` possible outcomesNow, count how many times we can have an Even number of dots and Heads
This means there are `3` times that we can have an Even number of dots and HeadsFinally, solve for the probabilityfavourable outcomes`=``3`total outcomes`=``12`$$ \mathsf{P(Even\:and\:Heads)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{3}}{\color{#007DDC}{12}}$$ Substitute values `=` $$\frac{1}{4}$$ `1/4` -
Question 6 of 8
6. Question
Find the probability of tossing `2` coins and getting:`(a) 2` Tails`(b)` Heads and TailsWrite fractions in the format “a/b”-
`(a)` (1/4, ¼)`(b)` (½, 1/2, 2/4)
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 getting `2` Tails.Set up a lattice showing all possible results for tossing `2` coins
This means there are `4` possible outcomesNow, count how many times we can have `2` Tails
This means there is `1` time that we can have `2` TailsFinally, find the probability of getting `2` Tailsfavourable outcomes`=``1`total outcomes`=``4`$$ \mathsf{P(2\:Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{4}}$$ Substitute values `(b)` Find the probability of getting Heads and Tails.Use the lattice from part `(a)` and count how many times we can have Heads and Tails
This means there are `2` times that we can have Heads and TailsFinally, find the probability of getting Heads and Tailsfavourable outcomes`=``2`total outcomes`=``4`$$ \mathsf{P(Heads\:and\:Tails)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{2}}{\color{#007DDC}{4}}$$ Substitute values `=` $$\frac{1}{2}$$ `(a) 1/4``(b) 1/2` -
-
Question 7 of 8
7. Question
Find the probability of rolling `2` normal six-sided dice and getting:`(a)` Sum of `9` or `10``(b)` Sum of `6` or `12`Write fractions in the format “a/b”-
`(a)` (7/36)`(b)` (1/6, 6/36)
Hint
Help VideoCorrect
Correct!
Incorrect
Addition Rule (Mutually Exclusive Events)
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$`(a)` Find the probability of getting a Sum of `9` or `10`.Set up a lattice showing all possible sums for the two dice
This means there are `36` possible outcomesNow, count how many times we can have a sum of `9` or `10`
Having a sum of `9` appears `4` timeHaving a sum of `10` appears `3` timesFinally, find the probability of getting a sum of `9` or `10` using Addition Rulefavourable outcomes`=``4+3=7`total outcomes`=``36`$$ \mathsf{P(sum\:of\:9\:or\:10)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{7}}{\color{#007DDC}{36}}$$ Substitute values `(b)` Find the probability of getting a Sum of `6` or `12`.Use the lattice from part `(a)` and count how many times we can have a sum of `6` or `12`
Having a sum of `6` appears `5` timesHaving a sum of `12` appears `1` timesFinally, find the probability of getting a sum of `6` or `12` using Addition Rulefavourable outcomes`=``5+1=6`total outcomes`=``36`$$ \mathsf{P(sum\:of\:6\:or\:12)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{6}}{\color{#007DDC}{36}}$$ Substitute values `=` $$\frac{1}{6}$$ `(a) 7/36``(b) 1/6` -
-
Question 8 of 8
8. Question
Find the probability of rolling `2` normal six-sided dice and getting `1` or `5` dots on the diceWrite fractions in the format “a/b”- (5/9, 20/36)
Hint
Help VideoCorrect
Nice Job!
Incorrect
Addition Rule (Mutually Exclusive Events)
$$\mathsf{P(A\:or\:B)}=\mathsf{P(A)}+\mathsf{P(B)}$$Events are mutually exclusive if they cannot
occur simultaneously.Probability Formula
$$\mathsf{P(E)}=\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$Set up a lattice showing all possible results for rolling two dice
This means there are `36` possible outcomesNow, count how many times `1` or `5` dots will appear
`1` dot appear `11` times`5` dots appear `11` timesRemember that each roll must only be counted once.`2` rolls are counted twice since they result to having both `1` AND `5` dots.Hence, subtract `2/36` from the final probability.Solve for the final probability using Addition Rulefavourable outcomes`=``11+11=22`total outcomes`=``36`$$ \mathsf{P(1\:or\:5)} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}-\frac{2}{36}$$ Probability Formula `=` $$\frac{\color{#e65021}{22}}{\color{#007DDC}{36}}-\frac{2}{36}$$ Substitute values `=` $$\frac{20}{36}$$ `=` $$\frac{5}{9}$$ `5/9`