7.SP.C.8 math practice. Learn by doing.

7.SP.C.8 practice covers find probabilities of compound events using organized lists, tables, tree diagrams, and simulation. Work through 8 free questions with answers and explanations, then continue in the related math games.

Grade 72 mapped skills
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8 free questions · 0/0 correct

Practice playeasy

A meal has 3 mains and 4 sides. How many different (main, side) combinations are possible?

By the multiplication principle: 3×4=123 \times 4 = 12.

Practice playeasy

Two fair coins are flipped. What is the probability that both coins show heads?

P(heads on one coin)=12P(\text{heads on one coin}) = \dfrac{1}{2}. For two independent flips, P(both heads)=12×12=14P(\text{both heads}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}.

Practice playeasy

For two independent events, the probability of flipping two heads in a row: P(A)=12P(A)=\tfrac{1}{2}, P(B)=12P(B)=\tfrac{1}{2}. Find P(A and B)P(A \text{ and } B) as a simplified fraction.

Multiply: P(A and B)=12×12=14P(A \text{ and } B) = \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}.

Practice playeasy

A game spinner has 88 equal sections, exactly 22 of which are blue. A fair coin is then flipped. What is the probability of landing on blue on the spinner and getting tails on the coin?

P(blue)=28=14P(\text{blue}) = \dfrac{2}{8} = \dfrac{1}{4} and P(tails)=12P(\text{tails}) = \dfrac{1}{2}. So P(both)=14×12=18P(\text{both}) = \dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8}.

Practice playeasy

Two fair coins are flipped. How many outcomes are in the sample space?

Each coin has 22 outcomes, so the sample space has 2×2=42 \times 2 = 4 outcomes: HH, HT, TH, TT.

Practice playeasy

A standard six-sided number cube is rolled twice. What is the probability of rolling a 22 both times?

P(2 on one roll)=16P(2\text{ on one roll}) = \dfrac{1}{6}. For two independent rolls, P(both 2s)=16×16=136P(\text{both }2\text{s}) = \dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}.

Practice playeasy

For two independent events, the probability of rolling two 6's: P(A)=16P(A)=\tfrac{1}{6}, P(B)=16P(B)=\tfrac{1}{6}. Find P(A and B)P(A \text{ and } B) as a simplified fraction.

Multiply: P(A and B)=16×16=136P(A \text{ and } B) = \tfrac{1}{6} \times \tfrac{1}{6} = \tfrac{1}{36}.

Practice playeasy

The weather report says there is a 35\dfrac{3}{5} chance of no rain on Saturday and an independent 35\dfrac{3}{5} chance of no rain on Sunday. What is the probability that it does not rain on both days?

Multiply the independent probabilities: 35×35=925\dfrac{3}{5} \times \dfrac{3}{5} = \dfrac{9}{25}.

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Practice 7.SP.C.8 inside a game.

The placement test chooses the right difficulty while every match keeps the standard in play.