Two unbiased dice, each marked 1 to 6 on their faces, are rolled simultaneously. What is the probability that the sum of the outcomes on the top faces of both dice is 7 or 10 ?
- (a)0·15
- (b)0·20
- (c)0·25
- (d)0·30
Answer
Why
Correct — C, (c) 0·25.
The sample space. Two distinguishable dice, each with six faces, give 6 × 6 = 36 equally likely outcomes. Treating the dice as distinguishable is essential: (2, 5) and (5, 2) are two different outcomes, and counting them as one is the commonest error in this whole topic.
Outcomes with a sum of 7. There are six: (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1). Seven is the most likely total on two dice for exactly this reason — every value of the first die can be completed to seven by some value of the second.
Outcomes with a sum of 10. There are three: (4,6), (5,5) and (6,4). Note that (5,5) occurs only once, since both dice must show five; only the unequal pairs come in two orders.
Combining them. A sum cannot be both 7 and 10 at once, so the two events are MUTUALLY EXCLUSIVE and their counts simply add: 6 + 3 = 9 favourable outcomes.
The probability is 9/36 = 1/4 = 0·25.
A useful cross-check: for a sum s from 8 to 12, the number of ways is 13 − s, so a sum of 10 gives 3; for a sum from 2 to 7, the number of ways is s − 1, so a sum of 7 gives 6. Those two little rules reproduce the whole distribution of totals on two dice — 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 ways for totals 2 to 12 — and they add to 36, which is the check that nothing has been missed.
A quick way to be sure of the counts without listing. For a sum of 7, every one of the six faces on the first die can be completed by exactly one face on the second, so there are six ways. For a sum of 10, the first die must show at least 4, leaving 4, 5 and 6 — three ways. Reasoning about how many values the FIRST die can take, and then checking that each admits exactly one completion, is faster and safer than writing out ordered pairs.
Why the others are wrong
- (a)0·15 — Close to 5/36, which is the probability of a sum of 8 or of 6, and also to what a candidate gets by counting the pairs for 10 as if unordered while counting those for 7 correctly. Every distinct ordered outcome must be counted, and for a sum of 10 those are (4,6), (5,5) and (6,4) — three, not two.
- (b)0·20 — 0·20 is 7·2 favourable outcomes out of 36, which no whole count can produce; the nearest genuine values are 7/36 and 8/36. It is the figure a candidate arrives at by estimating, or by counting six ways for the sum of seven and only one or two for the sum of ten. It is worth noticing that any probability in this problem must be a multiple of 1/36, so 0·20 could be discarded before any counting was done.
- (d)0·30 — Would need 10·8 favourable outcomes out of 36 — again not a whole number. It arises from over-counting, most often by adding a sum of 11 or a sum of 4 to the list, or by treating (5,5) as occurring twice. Each face pair with two equal values can occur in only one way, and that is the point (5,5) is in the question to test.
Concept
For equally likely outcomes, probability is the number of favourable outcomes divided by the total number of outcomes. With two dice the total is 36, and the counts for each sum form a symmetric triangle: 1 way for 2, 2 for 3, up to 6 for 7, then back down to 1 for 12. Because those counts add to 36, they provide their own check. Events that cannot occur together are mutually exclusive and their probabilities add directly; events that can occur together require the addition rule with the overlap subtracted, P(A or B) = P(A) + P(B) − P(A and B). Two dice are independent, so the probability of a particular ordered pair is 1/6 × 1/6 = 1/36, which is the same thing expressed multiplicatively.
Probability items in these papers are almost always dice, coins or a bag of coloured balls, and the discipline they test is careful enumeration. The examiner's standard devices are all present here: a sum with several representations, a sum containing a doublet that occurs only once, and a disjunction of two target sums so that the candidate must handle both. Options that are not multiples of 1/36 are a useful shortcut for eliminating wrong answers.
Probability items in this paper are enumeration exercises, and the examiner's devices are standard: a target sum with several representations, a doublet that occurs only once, and a disjunction so that two counts must be made rather than one. Each of those is a place where a count goes wrong. The structural check that catches most errors is that every probability over two dice must be a multiple of one thirty-sixth — a test that disposes of two options here before any counting begins, and one worth applying to every dice question in a few seconds.
Key facts
- Two dice give 36 equally likely ordered outcomes.
- A sum of 7 occurs in 6 ways — it is the most likely total.
- A sum of 10 occurs in 3 ways: (4,6), (5,5) and (6,4).
- A doublet such as (5,5) can occur in only one way, unlike unequal pairs, which occur in two.
- Sums of 7 and 10 are mutually exclusive, so their counts add: 6 + 3 = 9.
- The probability is 9/36 = 1/4 = 0·25.
- Number of ways for a sum s: s − 1 for s from 2 to 7, and 13 − s for s from 8 to 12.
- Every probability in a two-dice problem is a multiple of 1/36, which is a quick test of any offered option.
Study next
Common traps
- Treating the dice as indistinguishable and counting 21 outcomes instead of 36.
- Counting a doublet twice.
- Missing one of the two target sums in a disjunctive question.
- Accepting an answer that is not a multiple of 1/36.
Dice probability appears regularly in this paper's reasoning block. Write the number of ways for each required sum, add them if the events cannot coexist, and divide by 36 — and use the multiple-of-1/36 test to discard options before spending time on the count. Dice, coins and coloured balls supply almost every probability item in this family, and each is answered by counting the sample space first and the favourable cases second, in that order.
Related PYQs
EPFO_APFC_2023_Q74Suppose that x and y are distinct variables that take values from {1, 2, 3, 4, 5, 6}. What is the probability that the value of the expression xy + x + y is even?
- (a) 1/2
- (b) 1/3
- (c) 1/4
- (d) 1/5
Answer(d) 1/5
The probability item from the earlier paper, built on the same method — enumerate the admissible ordered pairs from a small set and divide by the size of the sample space.
Practice
- practice — not a real PYQ
Two unbiased dice are rolled together. What is the probability that the sum of the numbers on the top faces is 9 ?
- (a)1/12
- (b)1/9
- (c)1/6
- (d)5/36
Answer(b) 1/9
- practice — not a real PYQ
Two unbiased dice are rolled together. What is the probability that both dice show the same number ?
- (a)1/12
- (b)1/6
- (c)1/3
- (d)1/2
Answer(b) 1/6