Using Permutations and Combinations in Probability - Statistics
Card 1 of 30
State the addition rule for any two events $A$ and $B$.
State the addition rule for any two events $A$ and $B$.
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$P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Adds probabilities but subtracts overlap to avoid double-counting.
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Adds probabilities but subtracts overlap to avoid double-counting.
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A $5$-card hand is drawn from a $52$-card deck. What is $P(\text{all hearts})$?
A $5$-card hand is drawn from a $52$-card deck. What is $P(\text{all hearts})$?
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$\frac{\binom{13}{5}}{\binom{52}{5}}$. Choose 5 hearts from 13 divided by all ways to choose 5 from 52.
$\frac{\binom{13}{5}}{\binom{52}{5}}$. Choose 5 hearts from 13 divided by all ways to choose 5 from 52.
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State the formula for permutations of $n$ items taken $r$ at a time, $P(n,r)$.
State the formula for permutations of $n$ items taken $r$ at a time, $P(n,r)$.
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$P(n,r)=\frac{n!}{(n-r)!}$. Order matters, so divide $n!$ by $(n-r)!$ to remove unused items.
$P(n,r)=\frac{n!}{(n-r)!}$. Order matters, so divide $n!$ by $(n-r)!$ to remove unused items.
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State the formula for combinations of $n$ items taken $r$ at a time, $\binom{n}{r}$.
State the formula for combinations of $n$ items taken $r$ at a time, $\binom{n}{r}$.
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$\binom{n}{r}=\frac{n!}{r!(n-r)!}$. Order doesn't matter, so divide by $r!$ to remove duplicate arrangements.
$\binom{n}{r}=\frac{n!}{r!(n-r)!}$. Order doesn't matter, so divide by $r!$ to remove duplicate arrangements.
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What is the relationship between permutations and combinations: express $P(n,r)$ using $\binom{n}{r}$?
What is the relationship between permutations and combinations: express $P(n,r)$ using $\binom{n}{r}$?
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$P(n,r)=\binom{n}{r}r!$. Each combination has $r!$ permutations when order is considered.
$P(n,r)=\binom{n}{r}r!$. Each combination has $r!$ permutations when order is considered.
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Identify whether order matters for a permutation versus a combination.
Identify whether order matters for a permutation versus a combination.
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Permutation: order matters; combination: order does not. Key distinction: ABC ≠ BAC for permutations, but ABC = BAC for combinations.
Permutation: order matters; combination: order does not. Key distinction: ABC ≠ BAC for permutations, but ABC = BAC for combinations.
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State the formula for the number of distinct permutations of $n$ items with repeats $n_1,n_2,\dots,n_k$.
State the formula for the number of distinct permutations of $n$ items with repeats $n_1,n_2,\dots,n_k$.
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$\frac{n!}{n_1!n_2!\cdots n_k!}$. Divides $n!$ by factorials of repeat counts to eliminate duplicates.
$\frac{n!}{n_1!n_2!\cdots n_k!}$. Divides $n!$ by factorials of repeat counts to eliminate duplicates.
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State the multiplication rule for counting outcomes in a multi-stage process.
State the multiplication rule for counting outcomes in a multi-stage process.
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Total outcomes $=a_1a_2\cdots a_k$. Each stage multiplies the number of choices at that stage.
Total outcomes $=a_1a_2\cdots a_k$. Each stage multiplies the number of choices at that stage.
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State the classical probability formula using equally likely outcomes.
State the classical probability formula using equally likely outcomes.
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$P(E)=\frac{\text{favorable}}{\text{total}}$. Assumes all outcomes are equally likely to occur.
$P(E)=\frac{\text{favorable}}{\text{total}}$. Assumes all outcomes are equally likely to occur.
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State the complement rule for probability.
State the complement rule for probability.
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$P(E^c)=1-P(E)$. The probability of not-E equals 1 minus the probability of E.
$P(E^c)=1-P(E)$. The probability of not-E equals 1 minus the probability of E.
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State the probability rule for independent events $A$ and $B$.
State the probability rule for independent events $A$ and $B$.
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$P(A\cap B)=P(A)P(B)$. For independent events, multiply their individual probabilities.
$P(A\cap B)=P(A)P(B)$. For independent events, multiply their individual probabilities.
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Compute $P(7,3)$.
Compute $P(7,3)$.
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$210$. $P(7,3) = \frac{7!}{4!} = \frac{5040}{24} = 210$.
$210$. $P(7,3) = \frac{7!}{4!} = \frac{5040}{24} = 210$.
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Compute $\binom{7}{3}$.
Compute $\binom{7}{3}$.
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$35$. $\binom{7}{3} = \frac{7!}{3!4!} = \frac{5040}{6 \times 24} = 35$.
$35$. $\binom{7}{3} = \frac{7!}{3!4!} = \frac{5040}{6 \times 24} = 35$.
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How many distinct arrangements of the letters in $\text{LEVEL}$ are there?
How many distinct arrangements of the letters in $\text{LEVEL}$ are there?
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$\frac{5!}{2!2!}=30$. LEVEL has 5 letters with L and E each appearing twice.
$\frac{5!}{2!2!}=30$. LEVEL has 5 letters with L and E each appearing twice.
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A $5$-digit code uses digits $0$-$9$ with repetition allowed. How many codes are possible?
A $5$-digit code uses digits $0$-$9$ with repetition allowed. How many codes are possible?
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$10^5$. Each of 5 positions can be any of 10 digits: $10 \times 10 \times 10 \times 10 \times 10$.
$10^5$. Each of 5 positions can be any of 10 digits: $10 \times 10 \times 10 \times 10 \times 10$.
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A $5$-digit code uses digits $0$-$9$ with no repetition. How many codes are possible?
A $5$-digit code uses digits $0$-$9$ with no repetition. How many codes are possible?
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$P(10,5)=30240$. $P(10,5) = \frac{10!}{5!} = 10 \times 9 \times 8 \times 7 \times 6 = 30240$.
$P(10,5)=30240$. $P(10,5) = \frac{10!}{5!} = 10 \times 9 \times 8 \times 7 \times 6 = 30240$.
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From $10$ students, how many ways can you choose a $3$-person committee?
From $10$ students, how many ways can you choose a $3$-person committee?
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$\binom{10}{3}=120$. Order doesn't matter for committee selection.
$\binom{10}{3}=120$. Order doesn't matter for committee selection.
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From $10$ students, how many ways can you choose a president, vice president, and secretary?
From $10$ students, how many ways can you choose a president, vice president, and secretary?
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$P(10,3)=720$. Order matters since positions are distinct.
$P(10,3)=720$. Order matters since positions are distinct.
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Two cards are drawn without replacement. What is $P(\text{both aces})$?
Two cards are drawn without replacement. What is $P(\text{both aces})$?
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$\frac{\binom{4}{2}}{\binom{52}{2}}$. Choose 2 aces from 4 divided by all ways to choose 2 from 52.
$\frac{\binom{4}{2}}{\binom{52}{2}}$. Choose 2 aces from 4 divided by all ways to choose 2 from 52.
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A box has $6$ red and $4$ blue marbles. Two are drawn without replacement. What is $P(\text{one red and one blue})$?
A box has $6$ red and $4$ blue marbles. Two are drawn without replacement. What is $P(\text{one red and one blue})$?
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$\frac{\binom{6}{1}\binom{4}{1}}{\binom{10}{2}}=\frac{8}{15}$. Ways to pick 1 red and 1 blue divided by all ways to pick 2.
$\frac{\binom{6}{1}\binom{4}{1}}{\binom{10}{2}}=\frac{8}{15}$. Ways to pick 1 red and 1 blue divided by all ways to pick 2.
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Find the probability of drawing $2$ aces in $2$ draws without replacement from a $52$-card deck.
Find the probability of drawing $2$ aces in $2$ draws without replacement from a $52$-card deck.
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$\frac{C(4,2)}{C(52,2)}=\frac{1}{221}$. Choose 2 from 4 aces over choose 2 from 52 cards.
$\frac{C(4,2)}{C(52,2)}=\frac{1}{221}$. Choose 2 from 4 aces over choose 2 from 52 cards.
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Find the probability of getting exactly $2$ heads in $3$ fair coin flips.
Find the probability of getting exactly $2$ heads in $3$ fair coin flips.
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$\frac{C(3,2)}{2^3}=\frac{3}{8}$. Choose 2 heads positions from 3, divide by $2^3$ total outcomes.
$\frac{C(3,2)}{2^3}=\frac{3}{8}$. Choose 2 heads positions from 3, divide by $2^3$ total outcomes.
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Compute the number of $4$-digit codes using digits $0$-$9$ with no repetition allowed.
Compute the number of $4$-digit codes using digits $0$-$9$ with no repetition allowed.
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$P(10,4)=10\cdot 9\cdot 8\cdot 7=5040$. 10 choices, then 9, then 8, then 7 (no repeats).
$P(10,4)=10\cdot 9\cdot 8\cdot 7=5040$. 10 choices, then 9, then 8, then 7 (no repeats).
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State the formula for the permutation of $n$ objects taken $r$ at a time, $P(n,r)$.
State the formula for the permutation of $n$ objects taken $r$ at a time, $P(n,r)$.
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$P(n,r)=\frac{n!}{(n-r)!}$. Divides $n!$ by $(n-r)!$ to remove the unused $(n-r)$ objects.
$P(n,r)=\frac{n!}{(n-r)!}$. Divides $n!$ by $(n-r)!$ to remove the unused $(n-r)$ objects.
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Compute the number of $4$-digit codes using digits $0$-$9$ with repetition allowed.
Compute the number of $4$-digit codes using digits $0$-$9$ with repetition allowed.
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$10^4=10000$. Each position has 10 choices independently.
$10^4=10000$. Each position has 10 choices independently.
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State the formula for the combination of $n$ objects taken $r$ at a time, $C(n,r)$.
State the formula for the combination of $n$ objects taken $r$ at a time, $C(n,r)$.
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$C(n,r)=\frac{n!}{r!(n-r)!}$. Divides by $r!$ to remove order from permutations.
$C(n,r)=\frac{n!}{r!(n-r)!}$. Divides by $r!$ to remove order from permutations.
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What is the fundamental counting principle for $k$ stages with $n_1,\dots,n_k$ choices?
What is the fundamental counting principle for $k$ stages with $n_1,\dots,n_k$ choices?
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$n_1\cdot n_2\cdots n_k$. Multiply choices at each stage for total outcomes.
$n_1\cdot n_2\cdots n_k$. Multiply choices at each stage for total outcomes.
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What is the probability formula using equally likely outcomes and counting?
What is the probability formula using equally likely outcomes and counting?
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$P(A)=\frac{#A}{#S}$. Favorable outcomes divided by total outcomes.
$P(A)=\frac{#A}{#S}$. Favorable outcomes divided by total outcomes.
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Compute the number of distinct permutations of the letters in $\text{MISSISSIPPI}$.
Compute the number of distinct permutations of the letters in $\text{MISSISSIPPI}$.
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$\frac{11!}{4!4!2!}=34650$. Divide by factorials of repeated letters: M(1), I(4), S(4), P(2).
$\frac{11!}{4!4!2!}=34650$. Divide by factorials of repeated letters: M(1), I(4), S(4), P(2).
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Compute $C(8,3)$, the number of unordered selections of $3$ from $8$.
Compute $C(8,3)$, the number of unordered selections of $3$ from $8$.
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$C(8,3)=56$. $\frac{8!}{3!5!}=\frac{336}{6}=56$.
$C(8,3)=56$. $\frac{8!}{3!5!}=\frac{336}{6}=56$.
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