Comparing Probabilities - ISEE Upper Level: Mathematics Achievement
Card 1 of 20
Which is larger: $
$
$P=
\frac{3}{8}$ or $P=\frac{2}{5}$?
Which is larger: $ $ $P= \frac{3}{8}$ or $P=\frac{2}{5}$?
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$\frac{2}{5}$. Comparing $\frac{3}{8}=0.375$ and $\frac{2}{5}=0.4$ shows $\frac{2}{5}$ is larger.
$\frac{2}{5}$. Comparing $\frac{3}{8}=0.375$ and $\frac{2}{5}=0.4$ shows $\frac{2}{5}$ is larger.
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What rule lets you compare two probabilities by converting each to a common denominator?
What rule lets you compare two probabilities by converting each to a common denominator?
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Rewrite each as an equivalent fraction with the same denominator, then compare numerators. Converting fractions to a common denominator equalizes the scale, allowing direct numerator comparison to determine which probability is greater.
Rewrite each as an equivalent fraction with the same denominator, then compare numerators. Converting fractions to a common denominator equalizes the scale, allowing direct numerator comparison to determine which probability is greater.
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What is the rule for comparing probabilities given as decimals?
What is the rule for comparing probabilities given as decimals?
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The larger decimal represents the larger probability. Decimals express probabilities on a uniform scale from 0 to 1, so numerical comparison directly identifies the greater value.
The larger decimal represents the larger probability. Decimals express probabilities on a uniform scale from 0 to 1, so numerical comparison directly identifies the greater value.
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What is the rule for comparing probabilities given as percents?
What is the rule for comparing probabilities given as percents?
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The larger percent represents the larger probability. Percents represent probabilities scaled to 100, so the higher percent corresponds to the greater probability when compared directly.
The larger percent represents the larger probability. Percents represent probabilities scaled to 100, so the higher percent corresponds to the greater probability when compared directly.
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What is the probability form of odds $a:b$ in favor of an event?
What is the probability form of odds $a:b$ in favor of an event?
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$\frac{a}{a+b}$. Odds $a:b$ in favor mean the probability is the ratio of favorable to total outcomes.
$\frac{a}{a+b}$. Odds $a:b$ in favor mean the probability is the ratio of favorable to total outcomes.
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Which is more likely: drawing a red from $3$ red and $5$ blue, or drawing a head on a fair coin?
Which is more likely: drawing a red from $3$ red and $5$ blue, or drawing a head on a fair coin?
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Head on a fair coin. $P(\text{red})=\frac{3}{8}=0.375 < 0.5 = P(\text{head})$, so head has higher probability.
Head on a fair coin. $P(\text{red})=\frac{3}{8}=0.375 < 0.5 = P(\text{head})$, so head has higher probability.
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Which is larger: $P(A)=\frac{1}{3}$ or $P(A\cap B)$ if $A$ and $B$ are independent with $P(B)=\frac{1}{2}$?
Which is larger: $P(A)=\frac{1}{3}$ or $P(A\cap B)$ if $A$ and $B$ are independent with $P(B)=\frac{1}{2}$?
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$P(A)=\frac{1}{3}$. $P(A\cap B)=\frac{1}{3} \cdot \frac{1}{2}=\frac{1}{6} < \frac{1}{3}$, so $P(A)$ is larger.
$P(A)=\frac{1}{3}$. $P(A\cap B)=\frac{1}{3} \cdot \frac{1}{2}=\frac{1}{6} < \frac{1}{3}$, so $P(A)$ is larger.
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Which is more likely: getting $2$ heads in $2$ fair coin tosses, or rolling a $6$ on a fair die?
Which is more likely: getting $2$ heads in $2$ fair coin tosses, or rolling a $6$ on a fair die?
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Getting $2$ heads in $2$ tosses. $P(2\text{ heads})=(\frac{1}{2})^2=\frac{1}{4} > \frac{1}{6}=P(6)$, so 2 heads is more likely.
Getting $2$ heads in $2$ tosses. $P(2\text{ heads})=(\frac{1}{2})^2=\frac{1}{4} > \frac{1}{6}=P(6)$, so 2 heads is more likely.
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Which is more likely: rolling a multiple of $3$ on a fair die, or rolling an even number on a fair die?
Which is more likely: rolling a multiple of $3$ on a fair die, or rolling an even number on a fair die?
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Rolling an even number. Multiples of 3 have $P=\frac{2}{6}=\frac{1}{3}$, while even numbers have $P=\frac{3}{6}=\frac{1}{2} > \frac{1}{3}$.
Rolling an even number. Multiples of 3 have $P=\frac{2}{6}=\frac{1}{3}$, while even numbers have $P=\frac{3}{6}=\frac{1}{2} > \frac{1}{3}$.
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What is the complement rule used to compare an event and its “not” event?
What is the complement rule used to compare an event and its “not” event?
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$P(A^c)=1-P(A)$. The complement rule relates the probability of an event to 1 minus its complement for comparison.
$P(A^c)=1-P(A)$. The complement rule relates the probability of an event to 1 minus its complement for comparison.
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Which outcome is most likely for a fair die: ${1,2}$, ${3,4,5}$, or ${6}$?
Which outcome is most likely for a fair die: ${1,2}$, ${3,4,5}$, or ${6}$?
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${3,4,5}$. $P({3,4,5})=\frac{3}{6}=\frac{1}{2} > \frac{2}{6}$ or $\frac{1}{6}$ for the others.
${3,4,5}$. $P({3,4,5})=\frac{3}{6}=\frac{1}{2} > \frac{2}{6}$ or $\frac{1}{6}$ for the others.
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Which is larger: $P=\frac{5}{12}$ or $P=\frac{7}{18}$?
Which is larger: $P=\frac{5}{12}$ or $P=\frac{7}{18}$?
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$\frac{5}{12}$. $\frac{5}{12} \approx 0.417 > \frac{7}{18} \approx 0.389$ when compared as decimals.
$\frac{5}{12}$. $\frac{5}{12} \approx 0.417 > \frac{7}{18} \approx 0.389$ when compared as decimals.
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Which is larger: $P(\text{at least one }6\text{ in }2\text{ die rolls})$ or $\frac{1}{4}$?
Which is larger: $P(\text{at least one }6\text{ in }2\text{ die rolls})$ or $\frac{1}{4}$?
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$P(\text{at least one }6\text{ in }2\text{ rolls})$. $P(\text{at least one }6)=1-(\frac{5}{6})^2=\frac{11}{36} \approx 0.306 > 0.25$.
$P(\text{at least one }6\text{ in }2\text{ rolls})$. $P(\text{at least one }6)=1-(\frac{5}{6})^2=\frac{11}{36} \approx 0.306 > 0.25$.
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Which is larger: $P(A)=\frac{7}{12}$ or $P(A^c)$?
Which is larger: $P(A)=\frac{7}{12}$ or $P(A^c)$?
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$P(A)=\frac{7}{12}$. $P(A^c)=1-\frac{7}{12}=\frac{5}{12} < \frac{7}{12}$, so $P(A)$ is larger.
$P(A)=\frac{7}{12}$. $P(A^c)=1-\frac{7}{12}=\frac{5}{12} < \frac{7}{12}$, so $P(A)$ is larger.
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If $P(A)=0.2$ and $P(B)=0.35$ and $A,B$ are disjoint, what is $P(A\cup B)$?
If $P(A)=0.2$ and $P(B)=0.35$ and $A,B$ are disjoint, what is $P(A\cup B)$?
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$0.55$. Since disjoint, $P(A\cup B)=0.2+0.35=0.55$ by the addition rule.
$0.55$. Since disjoint, $P(A\cup B)=0.2+0.35=0.55$ by the addition rule.
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Which is more likely: at least one head in $2$ fair tosses, or drawing a spade from a standard deck?
Which is more likely: at least one head in $2$ fair tosses, or drawing a spade from a standard deck?
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At least one head in $2$ tosses. $P(\text{at least one head})=1-(\frac{1}{2})^2=\frac{3}{4} > \frac{13}{52}=\frac{1}{4}$.
At least one head in $2$ tosses. $P(\text{at least one head})=1-(\frac{1}{2})^2=\frac{3}{4} > \frac{13}{52}=\frac{1}{4}$.
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If one spinner has $3$ equal red sectors out of $8$ and another has $2$ equal red sectors out of $5$, which has higher $P(\text{red})$?
If one spinner has $3$ equal red sectors out of $8$ and another has $2$ equal red sectors out of $5$, which has higher $P(\text{red})$?
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The spinner with $\frac{2}{5}$ red. $\frac{3}{8}=0.375 < \frac{2}{5}=0.4$, so the second spinner has higher probability.
The spinner with $\frac{2}{5}$ red. $\frac{3}{8}=0.375 < \frac{2}{5}=0.4$, so the second spinner has higher probability.
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Which is larger: $P=0.37$ or $P=\frac{3}{10}$?
Which is larger: $P=0.37$ or $P=\frac{3}{10}$?
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$0.37$. $\frac{3}{10}=0.3$, and 0.37 is greater than 0.3 when compared directly as decimals.
$0.37$. $\frac{3}{10}=0.3$, and 0.37 is greater than 0.3 when compared directly as decimals.
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Which is larger: $P=45%$ or $P=0.4$?
Which is larger: $P=45%$ or $P=0.4$?
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$45%$. $45%=0.45$, which is greater than 0.4 when compared as decimals.
$45%$. $45%=0.45$, which is greater than 0.4 when compared as decimals.
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Which is more likely: an event with odds $3:2$ in favor, or probability $\frac{3}{7}$?
Which is more likely: an event with odds $3:2$ in favor, or probability $\frac{3}{7}$?
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Odds $3:2$ in favor. Odds $3:2$ in favor give $P=\frac{3}{5}=0.6$, which exceeds $\frac{3}{7}\approx^0.429$.
Odds $3:2$ in favor. Odds $3:2$ in favor give $P=\frac{3}{5}=0.6$, which exceeds $\frac{3}{7}\approx^0.429$.
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