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07 Feb 2026, 10:06
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C
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Difficulty:
75%
(hard)
Question Stats:
37% (01:39) correct
63%
(01:37)
wrong
based on 54
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If two different days of June, which has 30 days, are chosen at random, what is the probability that they are consecutive days?
(A) 1/30
(B) 1/29
(C) 1/15
(D) 2/29
(E) 1/10
(A) 1/30
(B) 1/29
(C) 1/15
(D) 2/29
(E) 1/10
07 Feb 2026, 13:11
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This is a classic probability problem that tests your understanding of combinations and favorable outcomes. The key trap here is thinking about "order matters" when it doesn't.
Step 1: Calculate total possible outcomes
We're choosing 2 different days from 30 days. Since we don't care about the order (choosing Day 5 and Day 6 is the same as choosing Day 6 and Day 5), we use combinations:
Total outcomes = C(30, 2) = 30 × 29 / 2 = 435
Step 2: Count favorable outcomes (consecutive days)
Consecutive pairs in June are: (1,2), (2,3), (3,4)... (29,30)
That's exactly 29 consecutive pairs.
Step 3: Calculate probability
Probability = Favorable outcomes / Total outcomes = 29/435
Simplify: 29/435 = 29/(30 × 29/2) = 29/(15 × 29) = 1/15
Answer: C
Common trap: Many students calculate this as 29/30 by thinking "pick any day, then the probability the next day is adjacent." That approach doesn't account for the fact that we're selecting two days simultaneously without replacement. The combination formula properly handles this.
Key concept tested: Probability with Combinations—when selecting without regard to order, always use C(n, r) for total outcomes.
Takeaway: When you see "chosen at random" with two items and no mention of order, think combinations first. Count your favorable outcomes carefully by listing patterns (like consecutive pairs here).
Step 1: Calculate total possible outcomes
We're choosing 2 different days from 30 days. Since we don't care about the order (choosing Day 5 and Day 6 is the same as choosing Day 6 and Day 5), we use combinations:
Total outcomes = C(30, 2) = 30 × 29 / 2 = 435
Step 2: Count favorable outcomes (consecutive days)
Consecutive pairs in June are: (1,2), (2,3), (3,4)... (29,30)
That's exactly 29 consecutive pairs.
Step 3: Calculate probability
Probability = Favorable outcomes / Total outcomes = 29/435
Simplify: 29/435 = 29/(30 × 29/2) = 29/(15 × 29) = 1/15
Answer: C
Common trap: Many students calculate this as 29/30 by thinking "pick any day, then the probability the next day is adjacent." That approach doesn't account for the fact that we're selecting two days simultaneously without replacement. The combination formula properly handles this.
Key concept tested: Probability with Combinations—when selecting without regard to order, always use C(n, r) for total outcomes.
Takeaway: When you see "chosen at random" with two items and no mention of order, think combinations first. Count your favorable outcomes carefully by listing patterns (like consecutive pairs here).
