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Question 1 of 5
1. Question
What is the probability of winning the lotto where there are `40` numbered balls and you are asked to pick `6` numbers, if you purchased only `1` entry?Hint
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Combination Formula
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$Probability Formula
$$ \mathsf{P=\frac{\color{#e65021}{favourable\:outcomes}}{\color{#007DDC}{total\:outcomes}}} $$Use the combinations formula to find the number of ways an item can be chosen `(r)` from the total number of items `(n)`.Remember that order is not important in Combinations.Find the different ways to draw `6` numbers `(r)` out of `40` total numbers `(n)``r=6``n=40`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{40}C_{\color{green}{6}}$$ `=` $$\frac{\color{purple}{40}!}{(\color{purple}{40}-\color{green}{6})!\color{green}{6}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{40!}{34! 6!}$$ `=` `3 838 380` Use the calculator’s factorial function for large numbers Now, find the probability that your `6` numbers will be chosen for the lottofavourable outcomes`=``1` (you have only purchased `1` entry)total outcomes`=``3 838 380` (total `6`-number combinations from `40` numbers)$$ \mathsf{P} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{3\;838\;380}}$$ Substitute values `1/(3 838 380)` -
Question 2 of 5
2. Question
What is the probability of winning the lotto where there are `40` numbered balls and you are asked to pick `6` numbers, if you purchased `190` entries?Hint
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Combination Formula
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$Probability Formula
$$ \mathsf{P=\frac{\color{#e65021}{favourable\:outcomes}}{\color{#007DDC}{total\:outcomes}}} $$Use the combinations formula to find the number of ways an item can be chosen `(r)` from the total number of items `(n)`.Remember that order is not important in Combinations.Find the different ways to draw `6` numbers `(r)` out of `40` total numbers `(n)``r=6``n=40`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{40}C_{\color{green}{6}}$$ `=` $$\frac{\color{purple}{40}!}{(\color{purple}{40}-\color{green}{6})!\color{green}{6}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{40!}{34! 6!}$$ `=` `3 838 380` Use the calculator’s factorial function for large numbers Now, find the probability that one of your entries will be chosen for the lottofavourable outcomes`=``190` (`190` entries)total outcomes`=``3 838 380` (total `6`-number combinations from `40` numbers)$$ \mathsf{P} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{190}}{\color{#007DDC}{3\;838\;380}}$$ Substitute values `=` $$\frac{1}{20\;202}$$ `1/(20 202)` -
Question 3 of 5
3. Question
In how many ways can `4` Jacks be chosen from a standard deck of `52` cards?Hint
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Combination Formula
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$Probability Formula
$$ \mathsf{P=\frac{\color{#e65021}{favourable\:outcomes}}{\color{#007DDC}{total\:outcomes}}} $$Use the combinations formula to find the number of ways an item can be chosen `(r)` from the total number of items `(n)`.Remember that order is not important in Combinations.First, find the different ways that `4` cards `(r)` can be chosen from a standard deck of `52` cards `(n)``r=4``n=52`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{52}C_{\color{green}{4}}$$ `=` $$\frac{\color{purple}{52}!}{(\color{purple}{52}-\color{green}{4})!\color{green}{4}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{52!}{48! 4!} $$ `=` `270 725` Use the calculator’s factorial function for large numbers Next, find the different ways that `4` Jacks `(r)` can be chosen from `4` Jacks `(n)` in the deck`r=4``n=4`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{4}C_{\color{green}{4}}$$ `=` $$\frac{\color{purple}{4}!}{(\color{purple}{4}-\color{green}{4})!\color{green}{4}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{4!}{0! 4!} $$ `=` $$ \frac{4!}{4!} $$ `0! =1` `=` `1` Now, find the probability of choosing `4` Jacks from a standard deck of cardsfavourable outcomes`=``1` (there is only `1` combination of Jacks)total outcomes`=``270 725` (total ways `4` cards can be drawn)$$ \mathsf{P} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{270\;725}}$$ Substitute values `1/(270 725)` -
Question 4 of 5
4. Question
What is the probability of drawing `4` Kings and `1` other card from a standard deck of `52` cards?Hint
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Combination Formula
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$Probability Formula
$$ \mathsf{P=\frac{\color{#e65021}{favourable\:outcomes}}{\color{#007DDC}{total\:outcomes}}} $$Use the combinations formula to find the number of ways an item can be chosen `(r)` from the total number of items `(n)`.Remember that order is not important in Combinations.First, find the number of ways that `4` King cards can be drawn. There are only `4` King cards available, which means:`r=4``n=4`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{4}C_{\color{green}{4}}$$ `=` $$\frac{\color{purple}{4}!}{(\color{purple}{4}-\color{green}{4})!\color{green}{4}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{4!}{0! 4!}$$ `=` $$ \frac{4\cdot3\cdot2\cdot1}{4\cdot3\cdot2\cdot1}$$ `0! =1` `=` $$1$$ Keeping in mind that `4` cards have already been drawn, find the different ways that `1` other card `(r)` can be chosen from a total of `48` remaining cards `(n)``r=1``n=48`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{48}C_{\color{green}{1}}$$ `=` $$\frac{\color{purple}{48}!}{(\color{purple}{48}-\color{green}{1})!\color{green}{1}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{48!}{47! 1!} $$ `=` $$ \frac{48\cdot47!}{47!} $$ `=` `48` `(47!)/(47!)` cancels out Multiply the two solved combinationsNumber of ways the `4` King cards can be dealt`=1`Number of ways one other card can be dealt`=48``1*48` `=` `48` Hence, there are `48` ways of drawing `4` Kings and one other card from a standard deck of `52` cards. This is the favourable outcome.Now, find the number of ways that `5` cards can be drawn from a total of `52` cards`r=5``n=52`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{52}C_{\color{green}{5}}$$ `=` $$\frac{\color{purple}{52}!}{(\color{purple}{52}-\color{green}{5})!\color{green}{5}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{52!}{47! 5!}$$ `=` `2 598 960` This is the total outcomeFinally, find the probability of drawing `4` Kings and `1` other card from a standard deckfavourable outcomes`=``48` (`4` Kings and `1` other card)total outcomes`=``2 598 960` (total ways of drawing `5` cards)$$ \mathsf{P} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{48}}{\color{#007DDC}{2\;598\;960}}$$ Substitute values `=` $$\frac{1}{54\;145}$$ `1/(54 145)` -
Question 5 of 5
5. Question
What is the probability of drawing `3` Queens and `2` other cards from a standard deck of `52` cards?Hint
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Incorrect
Combination Formula
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$Probability Formula
$$ \mathsf{P=\frac{\color{#e65021}{favourable\:outcomes}}{\color{#007DDC}{total\:outcomes}}} $$Use the combinations formula to find the number of ways an item can be chosen `(r)` from the total number of items `(n)`.Remember that order is not important in Combinations.First, find the ways that the `3` Queens can be dealt. There are `4` Queens available, which means:`r=3``n=4`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{4}C_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{4}!}{(\color{purple}{4}-\color{green}{3})!\color{green}{3}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{4!}{1! 3!}$$ `=` $$ \frac{4\color{#CC0000}{\cdot3\cdot2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1}}$$ `0! =1` `=` $$4$$ Keeping in mind that the `4` Queens cannot be chosen again, find the different ways that `2` other cards `(r)` can be chosen from a total of `48` remaining cards `(n)``r=2``n=48`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{48}C_{\color{green}{2}}$$ `=` $$\frac{\color{purple}{48}!}{(\color{purple}{48}-\color{green}{2})!\color{green}{2}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{48!}{46! 2!} $$ `=` $$ \frac{48\cdot47\cdot46!}{46!\cdot2\cdot1} $$ `=` `2256/2` `(46!)/(46!)` cancels out `=` `1128` Multiply the two solved combinationsNumber of ways the `3` Queens can be dealt`=4`Number of ways two other cards can be dealt`=1128``4*1128` `=` `4512` Hence, there are `4512` ways of dealing `3` Queens and two other cards from a standard deck of `52` cards. This is the favourable outcomeNow, find the number of ways that `5` cards can be drawn from a total of `52` cards`r=5``n=52`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{52}C_{\color{green}{5}}$$ `=` $$\frac{\color{purple}{52}!}{(\color{purple}{52}-\color{green}{5})!\color{green}{5}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{52!}{47! 5!}$$ `=` `2 598 960` This is the total outcomeFinally, find the probability of drawing `3` Queens and `2` other cards from a standard deckfavourable outcomes`=``4512` (`3` Queens and `2` other cards)total outcomes`=``2 598 960` (total ways of drawing `5` cards)$$ \mathsf{P} $$ `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcomes}}}{\color{#007DDC}{\mathsf{total\:outcomes}}}$$ Probability Formula `=` $$\frac{\color{#e65021}{4512}}{\color{#007DDC}{2\;598\;960}}$$ Substitute values `=` $$\frac{1}{576}$$ `1/576`
Quizzes
- Factorial Notation
- Fundamental Counting Principle 1
- Fundamental Counting Principle 2
- Fundamental Counting Principle 3
- Combinations 1
- Combinations 2
- Combinations with Restrictions 1
- Combinations with Restrictions 2
- Combinations with Probability
- Basic Permutations 1
- Basic Permutations 2
- Basic Permutations 3
- Permutation Problems 1
- Permutation Problems 2
- Permutations with Repetitions 1
- Permutations with Repetitions 2
- Permutations with Restrictions 1
- Permutations with Restrictions 2
- Permutations with Restrictions 3
- Permutations with Restrictions 4