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How do you compute probabilities of single and combined events on the MCAS?

Compute theoretical and experimental probabilities, apply the addition rule for either-or events and the multiplication rule for independent events, and find complements.

A Grade 10 Math MCAS answer on probability: theoretical versus experimental probability, the complement rule, the addition rule for either-or events, and the multiplication rule for independent events.

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  1. What this topic is asking
  2. Theoretical and experimental probability
  3. The complement rule
  4. The addition rule (or)
  5. The multiplication rule (and, independent)
  6. Try this

What this topic is asking

The Statistics and Probability category includes probability (the S-CP standards). On the Grade 10 MCAS you compute theoretical and experimental probabilities, use the complement rule, apply the addition rule for either-or events, and use the multiplication rule for independent events. The decisions, when to add and when to multiply, are exactly the reasoning the test checks, so understanding the structure of the question matters most.

Theoretical and experimental probability

Probability measures how likely an event is, on a scale from 0 (impossible) to 1 (certain). For equally likely outcomes:

P(event)=number of favorable outcomestotal number of outcomes.P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}.

For a bag of 3 red, 5 blue, 2 green marbles, P(blue)=510=12P(\text{blue}) = \frac{5}{10} = \frac{1}{2}. This is theoretical probability, computed from the known model.

Experimental probability comes from actually performing trials: if 200 flips of a coin give 108 heads, the experimental P(heads)=108200=0.54P(\text{heads}) = \frac{108}{200} = 0.54. Over many trials, experimental probability tends toward the theoretical value (the law of large numbers), but in a small sample it can differ.

The complement rule

The complement of an event is its not-happening. Because every outcome either is or is not in the event:

P(not A)=1P(A).P(\text{not } A) = 1 - P(A).

This is often the fast route to "at least one" problems. If P(at least one defective)P(\text{at least one defective}) is wanted and P(none defective)=0.7P(\text{none defective}) = 0.7, then P(at least one)=10.7=0.3P(\text{at least one}) = 1 - 0.7 = 0.3. Computing the complement and subtracting is usually easier than adding many cases.

The addition rule (or)

For the probability that AA or BB happens, the addition rule is:

P(A or B)=P(A)+P(B)P(A and B).P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B).

The subtraction avoids double-counting outcomes in both events. When AA and BB are mutually exclusive (cannot happen together, so P(A and B)=0P(A \text{ and } B) = 0), it simplifies to P(A)+P(B)P(A) + P(B). Drawing a card that is "a king or a queen" uses the simple form (a card cannot be both); "a king or a heart" needs the subtraction (the king of hearts is in both).

The multiplication rule (and, independent)

For two independent events (one does not affect the other's probability), the probability that both happen is the product:

P(A and B)=P(A)×P(B).P(A \text{ and } B) = P(A) \times P(B).

Coin flips, spinner spins, and draws with replacement are independent. Draws without replacement are not, because removing one item changes the next probability. The MCAS keeps most Grade 10 multiplication-rule items to independent cases, but reading whether the first outcome changes the second is the key check.

Try this

Q1. A die is rolled. What is P(not a 6)P(\text{not a 6})?

  • Cue. 116=561 - \frac{1}{6} = \frac{5}{6}.

Q2. Two coins are flipped. What is P(both tails)P(\text{both tails})?

  • Cue. 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.

Exam-style practice questions

Practice questions written in the style of MA DESE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Grade 10 Math MCAS (style)1 marksSelected-response. A bag has 3 red, 5 blue, and 2 green marbles. What is the probability of drawing a blue marble? (A) 12\frac{1}{2} (B) 510\frac{5}{10} (C) 15\frac{1}{5} (D) 52\frac{5}{2}
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The correct answer is (A).

Probability is favorable outcomes over total outcomes. There are 5 blue marbles out of 3+5+2=103 + 5 + 2 = 10 total, so P(blue)=510=12P(\text{blue}) = \frac{5}{10} = \frac{1}{2}. Choice (B) is correct in value but not reduced; choice (A) is the simplified form. A probability should be given in simplest form unless told otherwise.

Grade 10 Math MCAS (style)2 marksShort-answer. A fair coin is flipped twice. Find the probability of getting two heads, and explain why you multiply.
Show worked answer →

A 2-point item: one point for the answer, one for the independence reasoning.

Each flip is independent (one flip does not affect the other), so by the multiplication rule P(HH)=P(H)×P(H)=12×12=14P(\text{HH}) = P(H) \times P(H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. You multiply because the events are independent and you want both to happen. Listing the sample space HH, HT, TH, TT also shows 1 of 4 outcomes is two heads. Adding the probabilities instead of multiplying is the common error.

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