Conditional Probability and the Rules of Probability: Describing Events in Sample Spaces (CCSS.S-CP.1)
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Common Core High School Statistics And Probability › Conditional Probability and the Rules of Probability: Describing Events in Sample Spaces (CCSS.S-CP.1)
Two fair coins are flipped. The sample space is {HH, HT, TH, TT}. Let A be "at least one head" and B be "exactly one head." Which outcomes represent both events occurring?
{HH, HT, TH}
{HT, TH}
{HH}
{TT}
Explanation
Both events means the intersection $A \cap B$. Outcomes in $A$ are {HH, HT, TH}; outcomes in $B$ are {HT, TH}. Their intersection is {HT, TH}. Choice A is the union $A \cup B$, C and D are not in $B$.
A fair six-sided die is rolled. The sample space is {1, 2, 3, 4, 5, 6}. Let A be "even" and B be "greater than 3." Which outcomes represent at least one of the events?
{4, 6}
{2, 4, 6}
{4, 5, 6}
{2, 4, 5, 6}
Explanation
At least one means the union $A \cup B$. $A=\{2,4,6\}$ and $B=\{4,5,6\}$, so $A \cup B=\{2,4,5,6\}$. Choice A is $A \cap B$, B is just $A$, and C is just $B$.
One card is drawn at random from {2H, 5H, 8H, KH, 2C, 5C, 8C, KC}, where H = hearts and C = clubs. Let A be "the card is a heart." Which outcomes represent not A?
{2C, 5C, 8C, KC}
{2H, 5H, 8H, KH}
{5C, 8C, KC}
{2H, 2C, 5H, 5C}
Explanation
Not A means the complement $A^c$, i.e., all outcomes that are not hearts. That is {2C, 5C, 8C, KC}. Choice B is $A$ itself, C omits 2C, and D mixes hearts and clubs (not a complement).
Two fair six-sided dice are rolled: first die D1, second die D2. Let A be "the sum is 7" and B be "D1 shows an odd number." Which outcomes represent $A \cap B$?
{(1,6), (2,5), (3,4)}
{(1,6), (3,4), (5,2), (6,1)}
{(1,6), (3,4), (5,2)}
{(1,6), (2,5), (4,3), (6,1)}
Explanation
$A$ is all pairs summing to 7: {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}. Intersect with $B$ (D1 odd) gives {(1,6),(3,4),(5,2)}. Other choices include outcomes with D1 even or extra outcomes not in the intersection.
A fair six-sided die is rolled and a fair coin is flipped. Let A be "the die shows a multiple of 3" and B be "the coin shows heads." Which outcomes represent $A \cap B$?
{(3,H), (6,H), (1,H)}
{(3,H), (6,H)}
{(3,T), (6,T)}
{(1,H), (2,H), (3,H), (4,H), (5,H), (6,H)}
Explanation
$A=\{3,6\}$ on the die and $B=\{H\}$ on the coin. The intersection $A \cap B$ is the set of outcomes satisfying both: {(3,H), (6,H)}. Choice A adds an outcome not in $A$, C uses tails (not in $B$), and D is all heads outcomes ($B$ only).