Combinations with Probability

Question 1 of 5

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?

1. $1/(1 846)$
2. $1/(23 621 515)$
3. $1/(248 101)$
4. $1/(3 838 380)$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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 lotto

favourable 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

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?

1. $1/(2 582)$
2. $1/(1 202 202)$
3. $1/(20 202)$
4. $1/(202 202)$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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 lotto

favourable 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

In how many ways can

$4$ Jacks be chosen from a standard deck of

$52$ cards?

1. $1/(270 725)$
2. $1/(75 725)$
3. $1/(3 270 489)$
4. $1/(207 525)$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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 cards

favourable 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

What is the probability of drawing

$4$ Kings and

$1$ other card from a standard deck of

$52$ cards?

1. $1/(54 145)$
2. $1/(154 140)$
3. $1/(2 354 401)$
4. $3/(146 247)$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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 combinations

Number 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 outcome

Finally, find the probability of drawing

$4$ Kings and

$1$ other card from a standard deck

favourable 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

What is the probability of drawing

$3$ Queens and

$2$ other cards from a standard deck of

$52$ cards?

1. $1/(1576)$
2. $1/576$
3. $1/(280 481)$
4. $1/(28 155)$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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 combinations

Number 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 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 outcome

Finally, find the probability of drawing

$3$ Queens and

$2$ other cards from a standard deck

favourable 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

	$=$	$\frac{1}{576}$

$1/576$

Combinations with Probability

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Well Done!

Combination Formula

Probability Formula

2. Question

Hint

Correct!

Combination Formula

Probability Formula

3. Question

Hint

Fantastic!

Combination Formula

Probability Formula

4. Question

Hint

Nice Job!

Combination Formula

Probability Formula

5. Question

Hint

Great Work!

Combination Formula

Probability Formula

Quizzes