Question
Last week, Emily rode the bus to school on 3 of the 5 mornings and rode the bus home on 4 of the 5 afternoons. Based on last week’s events, what is the probability that Emily will ride the bus in the morning and the afternoon next Monday?
- A
- B
- C
- D
Show the answer
Correct answer: C
Using the joint probability formula, find the product of the two independent events. The probability that Emily will ride the bus to school on any given morning is . The probability that Emily will ride the bus home on any given afternoon is . Therefore, the probability that Emily will ride the bus in the morning and the afternoon is .
Question analysis
Last week Emily rode the bus to school on 3 of 5 mornings and home on 4 of 5 afternoons. This probability question asks how likely she is to ride the bus both ways next Monday, based on those frequencies. Students turn each count into a probability and combine the two, deciding how chances join when both events must happen.
- NYS standards
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- At grade 7 Compound probability is grade 7 work, often done with a table of outcomes. Each chance here comes from one week's frequencies, and joining the morning and afternoon events in one step is about the usual grade 7 demand.
- Estimated difficulty
- Moderate: several steps, or a close reading with one strong distractor.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Something you must already know moderate
- Turning words into math minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 5, 48 words.
- Corrected from the DOE handbook
- An inaccurate detail in the explanation was corrected.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Once the two chances are written, the question is how they combine. Pooling the week into 7 bus rides out of 10 trips, or adding fractions in some way, gives listed choices that look reasonable, since each lies between 0 and 1. Picturing the 25 pairings of a morning with an afternoon, 3 × 4 = 12 of which have both rides, shows why the chances multiply.
Something you must already know moderate Working it out
The rule that two independent events both happen with the product of their chances, 3/5 times 4/5, is not given, and “the morning and the afternoon” is the only signal for it. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The counts “3 of the 5 mornings” and “4 of the 5 afternoons” have to become two separate probabilities based on last week, not one pool of 10 trips. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Strategy: Turn each count into a probability, then multiply them, since both rides must happen; picturing a table of 5 mornings by 5 afternoons shows why multiplying works.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
The wrong answers, classified
A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Subtracts the morning chance from the afternoon chance, 4/5 − 3/5 = 1/5, an operation with no meaning for two rides that must both happen.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Adds the numerators, 3 + 4, but multiplies the denominators, 5 × 5, mixing the rule for adding fractions with the rule for multiplying them.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Pools the week’s trips into 7 bus rides out of 10, a rate for a single trip, as if riding both ways next Monday were just as likely.
What was corrected from the DOE handbook, and why
- Explanation: The DOE’s explanation says “The probability that Emily will ride the bus to school on any given afternoon is 4/5.” The question says she rode the bus home on 4 of the 5 afternoons; her rides to school are the morning ones. The site now says “ride the bus home on any given afternoon”.
Worked solution, step by step
Formulas used
- Probability from past results
- P is the probability, the chance something happens. You use what happened before to predict the chance it happens again. Emily rode in 3 of 5 mornings, so the chance for one morning is 3/5.
- Both of two independent events
- A and B are two events. Independent means one does not change the chance of the other. To find the chance that both happen, multiply their chances. The result is smaller than either one, since both must happen.
Sort out the question
Last week Emily:
- rode the bus to school on 3 of the 5 mornings;
- rode the bus home on 4 of the 5 afternoons.
You need the chance that she rides the bus in the morning and in the afternoon next Monday. The word and means both rides must happen.
Write each chance
Use last week to find each chance:
- Bus to school on a morning:
- Bus home on an afternoon:
Keep them apart. The morning rides go to school, and the afternoon rides go home.
Multiply the two chances
Both rides must happen, so multiply:
Why multiply? Pair each of the 5 mornings with each of the 5 afternoons. That makes 25 pairs. In of them, she rides both times. So the chance is 12 out of 25.
Avoid the tempting wrong choices
Watch out: pooling the week gives 7 rides out of 10 trips, or 7/10, choice D. That is the chance for one trip, not for two trips on one day.
Adding the tops but multiplying the bottoms gives 7/25, choice B. Subtracting the chances gives 1/5, choice A.
So the answer is C, .
Check your answer
The chance of both rides must be less than the chance of each ride alone. As decimals, 12/25 = 0.48. That is less than 3/5 = 0.6 and 4/5 = 0.8. Choice D, 7/10 = 0.7, is more than 3/5, so it cannot be right.
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The original DOE question
