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Chapter 13 · Probability

Exercise 13.1 · Q14: Two dice showing different numbers: sum 4

Exercise 13.1Conditional probability · Section 13.2
The question

Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Exam answer

Throwing two dice gives 6×6=366\times 6 = 36 equally likely outcomes (a,b)(a, b)⁠.

Let EE⁠: the sum is 4, and FF⁠: the two numbers are different. FF leaves out the 6 doubles (1,1),…,(6,6)(1, 1), \ldots, (6, 6)⁠, so

n(F)=36−6=30n(F) = 36 - 6 = 30 E={(1,3),(2,2),(3,1)}E = \{(1, 3), (2, 2), (3, 1)\}

so E∩F={(1,3),(3,1)}E\cap F = \{(1, 3), (3, 1)\}⁠.

By the definition of conditional probability,

P(E∣F)=P(E∩F)P(F)=236÷3036\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} \\ &= \frac{2}{36}\div\frac{30}{36} \end{aligned}

∴ P(E∣F)=115\therefore\ P(E|F) = \dfrac{1}{15}⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

"Given that the numbers are different" removes the 6 doubles from the 36 outcomes, leaving 30. Of the three ways to make a sum of 4, one is the double (2,2)(2, 2)⁠, which has been removed. So 2 of the 30 remaining outcomes have sum 4.

Given, to find, and the plan
Given
Two fair dice thrown together. Write an outcome as (a,b)(a, b)⁠, aa on the first die and bb on the second: 36 equally likely outcomes. It is known that a≠ba \ne b⁠.
To find
P(E∣F)P(E|F)⁠, where EE⁠: the sum is 4, and FF⁠: the two numbers are different.
Plan
(1) Count FF by leaving out the doubles. (2) List EE and keep the outcomes that are in FF⁠. (3) Divide.

Basics you need

1The 36 outcomes of two dice

The dice are thrown together but are different objects, so (1,3)(1, 3) and (3,1)(3, 1) are different outcomes: 6×6=366\times 6 = 36⁠. Basics: multiply the choices

2Counting by complement

"Different numbers" is everything except "the same number", so n(F)=n(S)−n(doubles)n(F) = n(S) - n(\text{doubles})⁠. Basics: events as sets

3Conditional probability

P(E∣F)=P(E∩F)P(F)P(E|F) = \dfrac{P(E\cap F)}{P(F)}⁠, provided P(F)≠0P(F) \ne 0 (Definition 1, Section 13.2). Lesson: conditional probability

Solution

  1. Count the given event FF

    n(F)=36−6=30,P(F)=3036.n(F) = 36 - 6 = 30,\qquad P(F) = \frac{30}{36}.
    The doubles (1,1),(2,2),…,(6,6)(1, 1), (2, 2), \ldots, (6, 6) are the 6 outcomes with equal numbers. Every other outcome shows different numbers. Counted directly: the first die has 6 choices and the second must differ from it, 5 choices, and 6×5=306\times 5 = 30⁠.
    Complement: the rest of the sample space. Basics: multiply the choices
    F: different numbers → 2 of 30 first die 1 1 2 2 3 3 4 4 5 5 6 6 second die 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 6 7 8 9 10 11 7 8 9 10 11 12 E ∩ F: sum 4, different F only E only: (2,2) not in F: doubles
    All 36 outcomes, first die across and second die down; each cell shows the sum. The dashed diagonal holds the 6 doubles, which are not in FF⁠. Of the three cells with sum 4, the double (2,2)(2, 2) is excluded; the two blue cells (1,3)(1, 3) and (3,1)(3, 1) remain, 2 of the 30 cells of FF⁠.

  2. List EE and intersect with FF

    E={(1,3),(2,2),(3,1)},E = \{(1, 3), (2, 2), (3, 1)\}, E∩F={(1,3),(3,1)},E\cap F = \{(1, 3), (3, 1)\}, P(E∩F)=236.P(E\cap F) = \frac{2}{36}.
    A sum of 4 needs a+b=4a + b = 4 with both between 1 and 6: a=1,2,3a = 1, 2, 3 with b=3,2,1b = 3, 2, 1⁠. The pair (2,2)(2, 2) shows the same number twice, so it is not in FF⁠.
    "And" is the intersection. Basics: events as sets

  3. Divide

    P(E∣F)=P(E∩F)P(F)=2/3630/36=236×3630=115.\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} = \frac{2/36}{30/36} \\ &= \frac{2}{36}\times\frac{36}{30} = \frac{1}{15}. \end{aligned}
    P(F)=3036≠0P(F) = \frac{30}{36} \ne 0⁠. The 36s cancel, leaving 230\frac{2}{30}⁠, and dividing top and bottom by 2 gives 115\frac{1}{15}⁠.
    Definition 1, Section 13.2. Basics: dividing by a fraction
Result

The probability that the sum is 4, given that the two numbers are different, is P(E∣F)=115P(E|F) = \dfrac{1}{15}⁠.

(Textbook answer: 115\frac{1}{15}⁠.)

Check it with numbers

Without the condition, a sum of 4 happens in 3 of the 36 outcomes: P(E)=336=112≈0.0833P(E) = \frac{3}{36} = \frac{1}{12} \approx 0.0833⁠. With it, P(E∣F)=230=115≈0.0667P(E|F) = \frac{2}{30} = \frac{1}{15} \approx 0.0667⁠. The chance falls because the condition removes one of the three ways to make 4, but only 6 of the 36 outcomes in all.

Common mistakes
  1. Keeping (2,2)(2, 2)⁠. It has sum 4, but its numbers are equal, so it is outside FF⁠. Counting it gives 330\frac{3}{30}⁠, which is wrong.
  2. Dividing by 36. 236\frac{2}{36} is P(E∩F)P(E\cap F)⁠. The given information shrinks the sample space to the 30 outcomes of FF⁠.
  3. Treating (1,3)(1, 3) and (3,1)(3, 1) as one outcome. The dice are distinct; with unordered pairs the outcomes would not be equally likely.
What to take away
  1. "Different numbers" on two dice: 36−6=3036 - 6 = 30 outcomes. "Same number": the 6 doubles.
  2. List the event, then strike out the outcomes that the condition rules out, and count what is left over n(F)n(F)⁠.

Every fact used, and where it comes from

FactUsed inSource
Two dice give 6×6=366\times 6 = 36 equally likely ordered outcomesGivenClass XI, Permutations and Combinations (Basics: multiply the choices)
n(F)=n(S)−n(complement)n(F) = n(S) - n(\text{complement})Step 1Class XI, Sets (Basics: events as sets)
P(E∣F)=P(E∩F)P(F)P(E|F) = \frac{P(E\cap F)}{P(F)}⁠, P(F)≠0P(F) \ne 0Step 3Section 13.2, Definition 1 (Lesson: conditional probability)
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