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Chapter 13 · Probability

Exercise 13.2 · Q16: Hostel newspapers: neither, and two conditionals

Exercise 13.2Conditional probability and the addition rule · Sections 13.2 and 13.4
The question

In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.

(a) Find the probability that she reads neither Hindi nor English newspapers.

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.

(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Exam answer

Let HH⁠: the student reads Hindi newspaper, and EE⁠: the student reads English newspaper.

P(H)=60100=35P(H) = \frac{60}{100} = \frac{3}{5} P(E)=40100=25P(E) = \frac{40}{100} = \frac{2}{5} P(H∩E)=20100=15P(H\cap E) = \frac{20}{100} = \frac{1}{5}

(a)

P(H′∩E′)=P((H∪E)′)[De Morgan]=1−P(H∪E)\begin{aligned} &P(H'\cap E') \\ &= P\big((H\cup E)'\big) \quad[\text{De Morgan}] \\ &= 1 - P(H\cup E) \end{aligned}

By the addition rule,

P(H∪E)=P(H)+P(E)−P(H∩E)=35+25−15=45\begin{aligned} &P(H\cup E) \\ &= P(H) + P(E) - P(H\cap E) \\ &= \frac{3}{5} + \frac{2}{5} - \frac{1}{5} = \frac{4}{5} \end{aligned} ∴ P(H′∩E′)=1−45\therefore\ P(H'\cap E') = 1 - \frac{4}{5}

P(reads neither) =15= \frac{1}{5}⁠.

(b)

P(E∣H)=P(E∩H)P(H)=1/53/5\begin{aligned} P(E|H) &= \frac{P(E\cap H)}{P(H)} \\ &= \frac{1/5}{3/5} \end{aligned}

P(E∣H)=13P(E|H) = \frac{1}{3}⁠.

(c)

P(H∣E)=P(H∩E)P(E)=1/52/5\begin{aligned} P(H|E) &= \frac{P(H\cap E)}{P(E)} \\ &= \frac{1/5}{2/5} \end{aligned}

∴ P(H∣E)=12\therefore\ P(H|E) = \dfrac{1}{2}⁠.

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

The one idea this problem is built on

Percentages of the students are probabilities for a student chosen at random. "Neither" is the complement of "at least one", found with the addition rule. "If she reads Hindi" is a condition: we look only at the Hindi readers, and ask what fraction of them also read English. That is P(E∣H)=P(E∩H)P(H)P(E|H) = \frac{P(E\cap H)}{P(H)}⁠.

Given, to find, and the plan
Given
60% read Hindi, 40% read English, 20% read both; one student is chosen at random.
To find
(a) P(neither), (b) P(English given Hindi), (c) P(Hindi given English).
Plan
(1) Name the events and turn the percentages into fractions. (2) (a) by De Morgan's law and the addition rule. (3) (b) and (c) by the definition of conditional probability. (4) Check by imagining 100 students.

Basics you need

1Percentages as probabilities

If 60% of the students read Hindi, a student picked at random reads Hindi with probability 60100\frac{60}{100}⁠: each student is equally likely to be picked. Basics: P(E) = n(E)/n(S)

2Neither, and the addition rule

"Neither" is H′∩E′=(H∪E)′H'\cap E' = (H\cup E)'⁠, and P(H∪E)=P(H)+P(E)−P(H∩E)P(H\cup E) = P(H) + P(E) - P(H\cap E)⁠. Basics: Venn diagrams

3Conditional probability

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

4Dividing fractions

1/53/5=15×53\frac{1/5}{3/5} = \frac15\times\frac53⁠. Basics: dividing by a fraction

Part (a): she reads neither newspaper

  1. Name the events and convert the percentages

    Let HH be the event "the student reads the Hindi newspaper" and EE the event "the student reads the English newspaper". Then

    P(H)=60100=35,P(E)=40100=25,P(H∩E)=20100=15.\begin{aligned} P(H) &= \frac{60}{100} = \frac{3}{5}, \\ P(E) &= \frac{40}{100} = \frac{2}{5}, \\ P(H\cap E) &= \frac{20}{100} = \frac{1}{5}. \end{aligned}
    The student is chosen at random, so the probability of each event is the fraction of students in it. Writing the percentages as fractions in lowest terms keeps every later step exact.
    P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}⁠. Basics: classical probability

  2. Neither is the complement of "at least one"

    P(H′∩E′)=P((H∪E)′)=1−P(H∪E).\begin{aligned} P(H'\cap E') &= P\big((H\cup E)'\big) \\ &= 1 - P(H\cup E). \end{aligned}

    By the addition rule,

    P(H∪E)=P(H)+P(E)−P(H∩E)=35+25−15=45.\begin{aligned} &P(H\cup E) \\ &= P(H) + P(E) - P(H\cap E) \\ &= \frac{3}{5} + \frac{2}{5} - \frac{1}{5} = \frac{4}{5}. \end{aligned}

    So

    P(H′∩E′)=1−45=15.P(H'\cap E') = 1 - \frac{4}{5} = \frac{1}{5}.
    H E S 2/5 1/5 1/5 neither: 1/5
    The four groups of students (not drawn to scale): Hindi only 35−15=25\frac35 - \frac15 = \frac25⁠, both 15\frac15⁠, English only 25−15=15\frac25 - \frac15 = \frac15⁠, and neither (shaded) 1−45=151 - \frac45 = \frac15⁠.
    A student reads neither paper exactly when she is outside H∪EH\cup E (De Morgan's law). Adding P(H)P(H) and P(E)P(E) counts the 20% who read both twice, so it is subtracted once.
    De Morgan's law and the addition rule (Class XI). Basics: the addition rule

Part (b): English, given Hindi

  1. Condition on the Hindi readers

    P(E∣H)=P(E∩H)P(H)=1/53/5=15×53=13.\begin{aligned} P(E|H) &= \frac{P(E\cap H)}{P(H)} \\ &= \frac{1/5}{3/5} = \frac{1}{5}\times\frac{5}{3} = \frac{1}{3}. \end{aligned}
    "If she reads Hindi newspaper" tells us the student is in HH⁠, so HH becomes the new sample space. Among the Hindi readers (35\frac35 of the students), those who also read English are 15\frac15 of the students, so the fraction is 15÷35\frac15\div\frac35⁠. P(H)=35≠0P(H) = \frac35 \ne 0⁠, so the definition applies. Note E∩H=H∩EE\cap H = H\cap E⁠.
    Definition 1, Section 13.2. Lesson: conditional probability

Part (c): Hindi, given English

  1. Condition on the English readers

    P(H∣E)=P(H∩E)P(E)=1/52/5=15×52=12.\begin{aligned} P(H|E) &= \frac{P(H\cap E)}{P(E)} \\ &= \frac{1/5}{2/5} = \frac{1}{5}\times\frac{5}{2} = \frac{1}{2}. \end{aligned}
    Now the condition is "she reads English", so we divide by P(E)=25≠0P(E) = \frac25 \ne 0⁠. The overlap is the same 15\frac15 as in Part (b), but it is a larger share of the smaller group: half the English readers also read Hindi, while only a third of the Hindi readers also read English.
    Definition 1, Section 13.2. Basics: dividing by a fraction
Result

(a) P(reads neither) =15= \frac15 (b) P(E∣H)=13P(E|H) = \frac13 (c) P(H∣E)=12P(H|E) = \frac12

(Textbook answer: (a) 15\frac15⁠, (b) 13\frac13⁠, (c) 12\frac12⁠.)

Check it with numbers

Imagine 100 students: 60 read Hindi, 40 read English, 20 read both. Then 60+40−20=8060 + 40 - 20 = 80 read at least one, and 100−80=20100 - 80 = 20 read neither: 20100=0.2\frac{20}{100} = 0.2⁠. Of the 60 Hindi readers, 20 read English: 2060≈0.3333\frac{20}{60} \approx 0.3333⁠. Of the 40 English readers, 20 read Hindi: 2040=0.5\frac{20}{40} = 0.5⁠.

Common mistakes
  1. Using 1−(0.6+0.4)=01 - (0.6 + 0.4) = 0 for (a). The 20% who read both are then counted twice.
  2. Mixing up (b) and (c). "If she reads Hindi" means divide by P(H)P(H)⁠; "if she reads English" means divide by P(E)P(E)⁠.
  3. Answering 15\frac15 for (b) or (c). That is P(H∩E)P(H\cap E)⁠, the share of all students who read both, not the share of one group.
What to take away
  1. "Neither" =1−P(at least one)= 1 - P(\text{at least one})⁠, with the addition rule for the union.
  2. P(E∣H)P(E|H) and P(H∣E)P(H|E) have the same numerator, P(E∩H)P(E\cap H)⁠, but different denominators.
  3. Here P(E∣H)=13P(E|H) = \frac13 differs from P(E)=25P(E) = \frac25⁠, so reading Hindi and reading English are not independent.

Every fact used, and where it comes from

FactUsed inSource
P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)} for a random choicePart (a), Step 1Class XI, Probability (recalled in Section 13.1)
De Morgan's law H′∩E′=(H∪E)′H'\cap E' = (H\cup E)'Part (a), Step 2Class XI, Sets
P(H∪E)=P(H)+P(E)−P(H∩E)P(H\cup E) = P(H) + P(E) - P(H\cap E)Part (a), Step 2Addition rule, Class XI (recalled in Section 13.1)
P(E∣F)=P(E∩F)P(F)P(E|F) = \frac{P(E\cap F)}{P(F)}⁠, P(F)≠0P(F) \ne 0Parts (b), (c)Definition 1, Section 13.2 (Lesson: conditional probability)
Independent only if P(E∣H)=P(E)P(E|H) = P(E)What to take awayDefinition 2, Section 13.4
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