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

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

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

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

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

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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 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

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1. Question

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Well Done!

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2. Question

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3. Question

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4. Question

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Nice Job!

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