The probability of drawing a red candy

2024 Handbook

2024 Test A Math · Question 17 of 57

Question

The probability of drawing a red candy at random from a bag of 25 candies is 25\frac{2}{5}. After 5 red candies are removed from the bag, what is the probability of randomly drawing a red candy from the bag?

  • A0
  • B110\frac{1}{10}
  • C15\frac{1}{5}
  • D14\frac{1}{4}
Show the answer

Correct answer: D

Start with a proportion to determine the total number of red candies in the bag:

25=r25\frac{2}{5} = \frac{r}{25}

25(25)=r\frac{2}{5}(25) = r

10=r10 = r

There are currently 10 red candies out of 25 candies in the bag. After 5 red candies are removed, there would be 5 red candies left in the bag and 20 total candies.

The probability of randomly drawing a red candy would then be 520=14\frac{5}{20} = \frac{1}{4}.

Question analysis

A probability question about a bag of 25 candies in which the chance of drawing red is 2/5; then 5 red candies are taken out. Students find the new probability of drawing a red candy. They must work out how many red candies there were, then rebuild the fraction from the counts left in the bag.

NYS standards
  • 6.SP.8a uniform probability models
  • 6.RP.3 ratio and rate reasoning
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Simple probability as a fraction of outcomes is grade 6 and 7 content. This adds a step back from the probability to a count and an update after removal, which is typical grade 7 multi-step work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 40 words.
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

The question gives a probability, not a count, so the first job is to rebuild the bag: 10 red candies among 25. Removing candies changes two numbers at once, and the strongest wrong choice comes from updating the red count while leaving the total at 25. A student who writes both new counts before forming the fraction avoids it.

  1. Turning words into math moderate Understanding the question

    The probability 2/5 has to be turned into a count, 10 red of 25, and "after 5 red candies are removed" lowers both the red count and the bag's total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three steps follow in order: 2/5 of 25 = 10, then 5 red among 20 candies left, then 5/20 reduced to lowest terms. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use 2/5 of 25 to find the number of red candies, subtract 5 from both the red count and the total, and write the new fraction in lowest terms.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=red candiesall candiesP = \frac{\text{red candies}}{\text{all candies}}
P is the probability, the chance of drawing a red candy. Red candies is how many red ones are in the bag. All candies is how many candies are in the bag in all. Each candy is equally likely, so the chance is the fraction of the bag that is red.
  1. Sort out the question

    The bag has 25 candies. The chance of drawing red is 25\frac{2}{5}. Then 5 red candies are taken out. You need the new chance of drawing red. Taking out candies changes both the red count and the total.

  2. Find the number of red candies

    A chance of 25\frac{2}{5} means 2 out of every 5 candies are red. Call the red count r.

    P=red candiesall candiesP = \frac{\text{red candies}}{\text{all candies}}

    25=r25\frac{2}{5} = \frac{r}{25}

    r=25×25=2×5=10r = \frac{2}{5} \times 25 = 2 \times 5 = 10

    There are 10 red candies. Watch out: do not stop at 25÷5=525 \div 5 = 5. You must also multiply by 2. Thinking there are 5 red candies leads to 0, choice A.

  3. Take out the 5 red candies

    red: 10−5=5\text{red: } 10 - 5 = 5

    all: 25−5=20\text{all: } 25 - 5 = 20

    Now 5 of the 20 candies are red. Watch out: the total goes down too. Keeping 25 gives 525=15\frac{5}{25} = \frac{1}{5}, choice C.

  4. Find the new probability

    P=520=14P = \frac{5}{20} = \frac{1}{4}

    To simplify, divide the top and bottom by 5. So the answer is D, 14\frac{1}{4}.

Check your answer

At first, 25\frac{2}{5} is 40% of the bag. Taking out only red candies should lower the chance. 14\frac{1}{4} is 25%, which is lower but not zero. That makes sense, because 5 red candies are still in the bag.

The original DOE question

Question 74 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 74. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).