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7th Grade · 7.SP.C.6

Experimental probability in 7th grade

Experimental probability is a chance found by trying something many times and counting what happens. In 7th grade, your child flips, rolls or spins, records the results and uses them to estimate a probability. Your child also learns the reverse: using a known probability to predict about how many times something will happen.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

7.SP.C.6 Approximate the probability of a chance event by collecting data on the chance process that produces it and observing its long-run relative frequency, and predict the approximate relative frequency given the probability.

What it means, in plain words

Each flip of a coin or roll of a number cube is called a trial. The relative frequency of an event is the number of times it happened divided by the number of trials. That fraction is the experimental probability of the event.

Theoretical probability is the chance worked out by reasoning, without doing the experiment. A fair coin has a theoretical probability of 12\frac{1}{2} for heads. Theoretical and experimental probability are rarely exactly equal. After 10 flips, heads might come up only 3 times. After 1,000 flips, the relative frequency of heads is usually very close to 0.5, so more trials give a better estimate.

The idea also works in reverse. If an event has a probability of 14\frac{1}{4}, your child predicts that in 200 trials it happens about 50 times. The word "about" matters: the real count will probably be close to 50, but not exactly 50.

Your child should be able to:

  • Find the relative frequency of an event from a table of results
  • Explain why more trials give a better estimate of a probability
  • Predict about how many times an event will happen, given its probability
  • Explain why a result close to the prediction, but not equal to it, is normal

Example

A paper cup is tossed 50 times and lands on its side 32 times. What is the experimental probability that it lands on its side?

  1. The event happened 32 times in 50 trials.
  2. Relative frequency: 3250=0.64\frac{32}{50} = 0.64.
  3. The experimental probability that the cup lands on its side is 0.64, or 64%.

Example

The probability of rolling a 1 on a number cube is 16\frac{1}{6}. If the cube is rolled 300 times, about how many times will it land on 1?

  1. Multiply the probability by the number of rolls: 16×300=50\frac{1}{6} \times 300 = 50.
  2. A count such as 46 or 55 would be normal.
  3. The cube will land on 1 about 50 times, but probably not exactly 50 times.

Where children usually get stuck

  • After a coin lands heads 6 times in 8 flips, your child decides the coin is unfair.

    A small number of trials can give a lopsided result by chance. The child treats 8 trials as if they settle the question.

    What helps. Flip a coin 8 times together, then 50 more times. Work out the relative frequency of heads after each round. It usually ends up closer to 0.5 after more flips.

  • Your child predicts exactly 50 ones in 300 rolls and says the experiment went wrong when the count is 46.

    The prediction was learned as an exact answer. Chance results land near a prediction and rarely on it.

    What helps. Use the word "about" in every prediction. Then ask whether 46 is close to 50. It is, so the result fits.

  • For a cup that landed on its side 32 times in 50 tosses, your child writes 3218\frac{32}{18}.

    The child divides by the other outcome, the 18 tosses that did not land on the side, instead of by all 50 trials.

    What helps. Ask how many tosses there were in all. That number always goes on the bottom of the fraction.

Try four practice questions

Real 7th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

  1. 1Warm-up
    In a simulation, an event has probability 0.50. What is the first trial checkpoint from which the running relative frequency stays within 0.10 of 0.50 for the rest of the table?
    Trial checkpointRunning relative frequency
    200.25
    400.45
    600.35
    800.52
    1000.47
    1200.55
    First checkpoint:trials
    Show the answer and how to solve it

    Answer. 80

    The answer is 80 because 80 is the first checkpoint where that row and all later rows stay between 0.40 and 0.60.

    1. 1The target is 0.50 and the tolerance is 0.10, so values must stay between 0.40 and 0.60.
    2. 2At 40 trials the value is 0.45, which is inside the band. But at 60 trials the value is 0.35, which is below 0.40 and outside the band. So 40 is too early.
    3. 3Starting at 80 trials, the remaining values are 0.52, 0.47, and 0.55. Every one of these is between 0.40 and 0.60, so 80 is the first checkpoint that works.

    Common mistake. A common mistake is choosing 40 because 0.45 is close to 0.50. But the 60-trial value of 0.35 falls outside the band, so 40 cannot be the answer.

    The idea underneath. Experimental probability does not have to match the target exactly; it can wiggle above and below it, as long as it stays inside the tolerance band.

  2. 2On level
    A spinner lands on blue with probability 30%. In 14 spins, predict how many land on blue. Round to the nearest whole number.
    Predicted count:
    Show the answer and how to solve it

    Answer. 4

    The answer is 4 because 30% of 14 is 4.2, and 4.2 rounds to 4.

    1. 1The chance of landing on blue is 30%, so you want 30% of the 14 spins. Convert the percent to a decimal.
    2. 2Multiply the probability by the number of spins.
    3. 3Round 4.2 to the nearest whole number. Since 0.2 is less than 0.5, it rounds down to 4.

    Common mistake. A common mistake is entering 30 because it is the percent, not the predicted number of blue spins. Another is rounding 4.2 up to 5, but 4.2 rounds down since 0.2 is less than 0.5.

    The idea underneath. Over many spins, a probability of 30% means about 3 out of every 10 spins land on blue. Multiply this share by 14 to predict the count. The actual count can differ.

  3. 3On level
    A student recorded the running relative frequency of rolling an even number in a number-cube simulation. Which claim best uses the table to estimate the long-run relative frequency?
    SpinsRunning relative frequency of even
    1000.62
    2000.56
    3000.53
    4000.51
    5000.50
    6000.51
    7000.49
    A
    The long-run estimate is about 0.50 because the later values level off near 0.50.
    B
    The long-run estimate is 0.62 because that is the first and greatest value in the table.
    C
    The long-run estimate is 0.49 because the last row is the most recent value.
    D
    No long-run estimate can be made yet because the values are not exactly the same.
    Show the answer and how to solve it

    The best estimate is about 0.50, because the last four values, 0.51, 0.50, 0.51, and 0.49, all hover around 0.50 — the claim that identifies this leveling-off pattern is correct.

    1. 1Look at how the running relative frequency changes as the number of spins increases.
    2. 2For a long-run estimate, focus on the rows near the bottom of the table. Early rows can shift a lot before settling down.
    3. 3The last four values are 0.51, 0.50, 0.51, and 0.49. They stay very close to 0.50, wobbling just above and below it.
    4. 4The claim that identifies about 0.50 as the center of the later cluster is the best estimate. The first-row value of 0.62 is not a good estimate because early rows can be far off. The last-row value of 0.49 alone is not the best estimate because the whole cluster—not just one row—should guide the estimate. And small differences among the later rows are normal; they do not prevent a long-run estimate.

    Common mistake. A common mistake is choosing 0.49 because it is the last value shown. The long-run estimate should come from the cluster of later values, not from one final row.

    The idea underneath. As more trials are added, the running relative frequency may still wiggle a little, so the long-run estimate is the center of the later wiggles, not any single row.

  4. 4Stretch
    A marble is drawn at random from a bag and then put back. The probability of drawing a red marble is 35%. If a marble is drawn 60 times, which range is the most reasonable for the number of times a red marble is drawn?
    A
    9 to 13
    B
    14 to 18
    C
    19 to 23
    D
    24 to 28
    Show the answer and how to solve it

    The answer is 19 to 23 because 35% of 60 is 21, and 21 falls in that range.

    1. 135% is a little more than one-third. One-third of 60 is 20, so the number of red marbles should be just above 20.
    2. 2The probability describes the long-run share: about 7 red-marble draws per 20 draws. With 3 groups of 20 in 60 draws, we expect about 21 red-marble draws. The actual count can vary.
    3. 321 falls in the range 19 to 23. The neighboring range 14 to 18 is centered at 16 (5 below), and the neighboring range 24 to 28 is centered at 26 (5 above).
    4. 4Since 21 is inside 19 to 23 and not inside any adjacent band, 19 to 23 is the most reasonable range.

    Common mistake. A common mistake is choosing 14 to 18 (slightly too low) or 24 to 28 (slightly too high) when the neighboring bands are close. Another mistake is choosing 9 to 13 by greatly underestimating 35% of 60.

    The idea underneath. A reasonable success count in repeated trials is centered near probability × number of trials, so choose the range whose center is closest to that product.