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Chapter 13 · Probability

Exercise 13.2 · Q12: Die tossed thrice: an odd number at least once

Exercise 13.2Independent events · Section 13.4
The question

A die is tossed thrice. Find the probability of getting an odd number at least once.

Exam answer

Let A1A_1⁠, A2A_2⁠, A3A_3 be the events of an odd number on the first, second and third toss.

P(Ai)=36=12,P(Ai′)=12P(A_i) = \frac{3}{6} = \frac{1}{2},\quad P(A_i') = \frac{1}{2}

The tosses are independent, so

P(no odd number)=P(A1′∩A2′∩A3′)=P(A1′) P(A2′) P(A3′)=12×12×12=18\begin{aligned} &P(\text{no odd number}) \\ &= P(A_1'\cap A_2'\cap A_3') \\ &= P(A_1')\,P(A_2')\,P(A_3') \\ &= \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2} = \frac{1}{8} \end{aligned}

"At least one odd" is the complement of "no odd number".

P(at least one odd)=1−P(no odd number)=1−18\begin{aligned} &P(\text{at least one odd}) \\ &= 1 - P(\text{no odd number}) \\ &= 1 - \frac{1}{8} \end{aligned}

∴\therefore Required probability =78= \dfrac{7}{8}⁠.

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

The one idea this problem is built on

"At least once" has many cases (odd on one toss, on two, on all three). Its complement has just one: no odd number at all, that is, even on every toss. The tosses do not affect each other, so that one case has probability 12×12×12\frac12\times\frac12\times\frac12⁠, and the answer is 1 minus it.

Given, to find, and the plan
Given
A fair die tossed three times. On each toss the faces 1 to 6 are equally likely, and the tosses do not affect one another.
To find
P(an odd number at least once).
Plan
(1) Name an event for each toss. (2) Turn "at least once" into its complement. (3) Multiply the three probabilities of "even", using independence. (4) Subtract from 1. (5) Check by counting all 63=2166^3 = 216 outcomes.

Basics you need

1Classical probability

P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)} for equally likely outcomes; a die has 3 odd faces (1, 3, 5) out of 6. Basics: P(E) = n(E)/n(S)

2The complement rule

P(E)=1−P(E′)P(E) = 1 - P(E')⁠. "At least one" is the complement of "none". Basics: events as sets

3Independent trials

Separate tosses of a die are independent experiments, so the probability that several of them each give a stated result is the product of the separate probabilities (Remarks (iii) and (iv), Section 13.4). Lesson: independent events

4Counting outcomes

Three tosses with 6 results each give 6×6×6=2166\times 6\times 6 = 216 outcomes. Basics: multiply the choices

Solution

  1. Name the events and find their probabilities

    Let A1A_1⁠, A2A_2⁠, A3A_3 be the events "an odd number on the first toss", "on the second toss" and "on the third toss". On any toss, 3 of the 6 faces are odd, so

    P(Ai)=36=12,P(Ai′)=1−12=12.\begin{aligned} P(A_i) &= \frac{3}{6} = \frac{1}{2}, \\ P(A_i') &= 1 - \frac{1}{2} = \frac{1}{2}. \end{aligned}

    Here Ai′A_i' is "an even number on toss ii".

    One event per toss lets us use what we know about each toss separately. "Not odd" on a die is "even" (2, 4 or 6), which is also 3 faces out of 6.
    P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)} and P(E′)=1−P(E)P(E') = 1 - P(E)⁠. Basics: classical probability

  2. Turn "at least once" into its complement

    "An odd number at least once" fails only when every toss is even, the event A1′∩A2′∩A3′A_1'\cap A_2'\cap A_3'⁠. So

    P(at least one odd)=1−P(A1′∩A2′∩A3′).\begin{aligned} &P(\text{at least one odd}) \\ &\quad = 1 - P(A_1'\cap A_2'\cap A_3'). \end{aligned}
    "At least once" covers 7 different patterns of odd and even (see the tree below), while its complement is a single pattern: even, even, even. One pattern is much easier to work out than seven.
    The complement rule. Basics: events as sets

  3. Multiply, because the tosses are independent

    P(A1′∩A2′∩A3′)=P(A1′) P(A2′) P(A3′)=12×12×12=18.\begin{aligned} &P(A_1'\cap A_2'\cap A_3') \\ &= P(A_1')\,P(A_2')\,P(A_3') \\ &= \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2} \\ &= \frac{1}{8}. \end{aligned}
    What happens on one toss does not change the chances on another, so the tosses are independent experiments (Remark (iii), Section 13.4), and the three events "even on toss ii" are mutually independent (Remark (iv)). For mutually independent events the probability that all of them happen is the product of their probabilities.
    Remarks (iii) and (iv), Section 13.4. Lesson: independent events

  4. Subtract from 1

    P(at least one odd)=1−18=78.\begin{aligned} P(\text{at least one odd}) &= 1 - \frac{1}{8} \\ &= \frac{7}{8}. \end{aligned}
    S O E O E O E O E O E O E O E OOO OOE OEO OEE EOO EOE EEO EEE: no odd, 1/8 every branch: 1/2 at least one odd: 7 paths, 7/8
    The tree of three tosses (O = odd, E = even); every branch has probability 12\frac12⁠, so every path has 18\frac18⁠. Seven paths contain at least one O. Only the path E, E, E (rose) has no odd number: 1−18=781 - \frac18 = \frac78⁠.
    "At least one odd" and "no odd" together cover every outcome without overlapping, so their probabilities add to 1. The tree shows the same thing: 7 of the 8 equally likely paths contain an O.
    Tree diagrams: multiply along a path, add paths. Basics: tree diagrams

  5. Check by counting all outcomes

    The three tosses give 63=2166^3 = 216 equally likely outcomes. "No odd number" means each toss shows 2, 4 or 6, which happens in 33=273^3 = 27 ways. So

    P(at least one odd)=216−27216=189216=78.\begin{aligned} P(\text{at least one odd}) &= \frac{216 - 27}{216} \\ &= \frac{189}{216} = \frac{7}{8}. \end{aligned}
    This route needs no independence argument, only counting, and it gives the same answer. 189=27×7189 = 27\times 7 and 216=27×8216 = 27\times 8⁠.
    Multiply the choices. Basics: counting
Result

P(an odd number at least once) =1−(12)3=78= 1 - \left(\frac12\right)^3 = \frac78⁠.

(Textbook answer: 78\frac78⁠.)

Check it with numbers

Of the 216 outcomes, 27 have no odd number and 189 have at least one: 189216=0.875\frac{189}{216} = 0.875 and 27216=0.125\frac{27}{216} = 0.125⁠, which add to 1. The seven patterns with at least one odd, each 0.1250.125⁠, also give 7×0.125=0.8757\times 0.125 = 0.875⁠.

Common mistakes
  1. Adding 12+12+12\frac12 + \frac12 + \frac12⁠. That gives 32\frac32⁠, more than 1. The events "odd on toss 1", "odd on toss 2", "odd on toss 3" overlap, so their probabilities cannot simply be added.
  2. Answering 18\frac18⁠. That is the probability of "no odd number", the complement; the question asks for "at least once".
  3. Answering 38\frac38⁠. That is "odd exactly once". "At least once" also includes twice and three times.
What to take away
  1. "At least one" questions: use 1−P(none)1 - P(\text{none})⁠.
  2. For independent trials, P(none)P(\text{none}) is a product of complements.
  3. For a fair die, odd and even each have probability 12\frac12⁠, so the tree has equal branches.

Every fact used, and where it comes from

FactUsed inSource
P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)} for equally likely outcomesSteps 1, 5Class XI, Probability (recalled in Section 13.1)
P(E′)=1−P(E)P(E') = 1 - P(E)Steps 1, 2, 4Class XI, Probability
Independent experiments: probabilities multiplyStep 3Remark (iii), Section 13.4 (Lesson: independent events)
Mutual independence of three eventsStep 3Remark (iv), Section 13.4
Odd numbers on two throws of a die are independentStep 3Example 11, Section 13.4 (Lesson: Examples 10 to 14)
Multiplication principle of countingStep 5Class XI, Permutations and Combinations
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