Class 12 Maths, Worked in Full Past papersGet updates
Chapter 13 · Probability

Miscellaneous · Q4: At most 6 of 10 people right-handed, P≈0.0128P \approx 0.0128

Miscellaneous ExerciseIndependent trials · Section 13.4
The question

Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Exam answer

Let pp = P(a person is right-handed) =90%=910= 90\% = \frac{9}{10}⁠, and q=1−p=110q = 1 - p = \frac{1}{10}⁠.

The 10 people are selected at random, so the 10 selections are independent trials. Let XX be the number of right-handed people in the sample.

For exactly rr right-handed people, the rr people can be chosen in 10Cr{}^{10}C_r ways. Each such choice has probability prq10−rp^r q^{10-r} (multiplication rule for independent events), and the choices are mutually exclusive. So

P(X=r)=10Cr prq10−rP(X = r) = {}^{10}C_r\, p^r q^{10-r}

Now P(X≤6)=1−P(X≥7)P(X \le 6) = 1 - P(X \ge 7)⁠, where

P(X≥7)=∑r=71010Cr(0.9)r(0.1)10−r=120⋅97+45⋅981010+10⋅99+9101010=573956280+19371024451010+3874204890+34867844011010=98720480161010\begin{aligned} &P(X \ge 7) \\ &= \sum_{r=7}^{10} {}^{10}C_r (0.9)^r (0.1)^{10-r} \\ &= \frac{120\cdot 9^7 + 45\cdot 9^8}{10^{10}} \\ &\quad + \frac{10\cdot 9^9 + 9^{10}}{10^{10}} \\ &= \frac{573956280 + 1937102445}{10^{10}} \\ &\quad + \frac{3874204890 + 3486784401}{10^{10}} \\ &= \frac{9872048016}{10^{10}} \end{aligned} P(X≤6)=1−98720480161010=1279519841010\begin{aligned} &P(X \le 6) \\ &= 1 - \frac{9872048016}{10^{10}} \\ &= \frac{127951984}{10^{10}} \end{aligned}

∴\therefore P(at most 6 are right-handed) =7996999625000000= \dfrac{7996999}{625000000} ≈0.0128\approx 0.0128⁠.

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

The one idea this problem is built on

Treat the 10 people as 10 independent trials, each "right-handed" with probability 910\frac{9}{10}⁠. Any one pattern of right- and left-handers has probability equal to a product, by the multiplication rule for independent events; the number of patterns with exactly rr right-handers is 10Cr{}^{10}C_r⁠. "At most 6" is unlikely when 9 in 10 people are right-handed, so it is quicker to find its complement, "7 or more", and subtract from 1.

Given, to find, and the plan
Given
90% of people are right-handed; a random sample of 10 people. Not written, but meant: each person in the sample is right-handed with probability 910\frac{9}{10}⁠, independently of the others (the population is large, so picking one person does not change the chances for the next).
To find
P(at most 6 of the 10 are right-handed), that is, P(0, 1, 2, 3, 4, 5 or 6 of them are).
Plan
(1) Set up the trials. (2) Use the complement "7 or more". (3) Find the probability of one pattern by the multiplication rule. (4) Count the patterns. (5) Add the four cases and subtract from 1.

Basics you need

1Independent events multiply

For mutually independent events, the probability that all of them happen is the product of their probabilities (Section 13.4, Remarks (iii) and (iv)). Lesson: independent events

2Choosing rr things from nn

The number of ways to choose which rr of nn people have a property is nCr=n!r! (n−r)!{}^{n}C_r = \frac{n!}{r!\,(n-r)!} (Class XI, Permutations and Combinations). Basics: counting

3Complement and addition

P(E′)=1−P(E)P(E') = 1 - P(E)⁠, and probabilities of events that cannot happen together add. Basics: rules of probability

4Percentages as fractions

90%=90100=910=0.990\% = \frac{90}{100} = \frac{9}{10} = 0.9⁠, so a person is left-handed with probability 110=0.1\frac{1}{10} = 0.1⁠. Basics: working with fractions

Solution

  1. Set up 10 independent trials

    Write R for a right-handed person and L for anyone else. For each person in the sample,

    P(R)=90%=910,P(L)=1−910=110.\begin{aligned} P(R) &= 90\% = \frac{9}{10}, \\ P(L) &= 1 - \frac{9}{10} = \frac{1}{10}. \end{aligned}

    The sample is random, so the 10 people are 10 independent trials. Let XX be the number of right-handed people among them; XX can be any of 0,1,…,100, 1, \ldots, 10⁠.

    Independence is what lets us multiply in Step 3. It is the natural reading of "a random sample" from a very large population: who the first person is does not change the chance that the second is right-handed.
    Independent experiments: probabilities multiply (Section 13.4, Remark (iii)). Lesson: independent events

  2. Use the complement

    "At most 6" means X=0,1,2,3,4,5X = 0, 1, 2, 3, 4, 5 or 66⁠: seven cases. Its complement, "at least 7", is only four cases, X=7,8,9,10X = 7, 8, 9, 10⁠. So

    P(X≤6)=1−P(X≥7).P(X \le 6) = 1 - P(X \ge 7).
    Every sample has some number of right-handers from 0 to 10, so "at most 6" and "at least 7" are complementary events. Four cases are less work than seven; Step 7 checks that the seven cases give the same answer.
    P(E′)=1−P(E)P(E') = 1 - P(E)⁠. Basics: rules of probability

  3. The probability of one pattern

    Take one particular pattern with exactly 7 right-handers, say the first 7 people are R and the last 3 are L: RRRRRRRLLL. By the multiplication rule for independent events,

    P(RRRRRRRLLL)=(910)7(110)3=971010.\begin{aligned} P(\text{RRRRRRRLLL}) &= \left(\frac{9}{10}\right)^7 \left(\frac{1}{10}\right)^3 \\ &= \frac{9^7}{10^{10}}. \end{aligned}

    Any other pattern with 7 R's and 3 L's, such as RLRRLRRRLR, has the same seven factors 910\frac{9}{10} and three factors 110\frac{1}{10}⁠, only in a different order, so it has the same probability 971010\frac{9^7}{10^{10}}⁠.

    Each of the 10 people is a separate independent event, so the probability that all ten turn out as the pattern says is the product of the ten separate probabilities. The order of the factors in a product does not matter.
    Mutually independent events multiply (Section 13.4, Remark (iv), extended to 10 events). Lesson: independent events

  4. Count the patterns, then add

    A pattern with exactly 7 R's is fixed by choosing which 7 of the 10 places hold an R. That can be done in

    10C7=10!7! 3!=120 ways.{}^{10}C_7 = \frac{10!}{7!\,3!} = 120 \text{ ways}.

    Two different patterns cannot both happen, so their probabilities add:

    P(X=7)=120×971010=120×47829691010=5739562801010.\begin{aligned} P(X = 7) &= 120 \times \frac{9^7}{10^{10}} \\ &= \frac{120 \times 4782969}{10^{10}} \\ &= \frac{573956280}{10^{10}}. \end{aligned}

    In the same way, for any rr from 0 to 10,

    P(X=r)=10Cr(910)r(110)10−r.P(X = r) = {}^{10}C_r \left(\frac{9}{10}\right)^r \left(\frac{1}{10}\right)^{10-r}.
    "Exactly 7 right-handers" is the union of the 120 patterns, and the patterns are mutually exclusive (a sample follows exactly one pattern). This general formula is the binomial distribution; here it comes straight from the multiplication rule and counting.
    nCr{}^{n}C_r counts the ways to choose rr places from nn⁠. Basics: counting

  5. The four cases 7, 8, 9, 10

    All four have the denominator 101010^{10}⁠:

    rr10Cr{}^{10}C_r10Cr 9r{}^{10}C_r\, 9^rP(X=r)P(X = r)
    7120573956280≈0.0574\approx 0.0574
    8451937102445≈0.1937\approx 0.1937
    9103874204890≈0.3874\approx 0.3874
    1013486784401≈0.3487\approx 0.3487

    Here 97=47829699^7 = 4782969⁠, 98=430467219^8 = 43046721⁠, 99=3874204899^9 = 387420489 and 910=34867844019^{10} = 3486784401⁠.

    For example, P(X=10)=(910)10P(X = 10) = \left(\frac{9}{10}\right)^{10}⁠: all ten right-handed, a single pattern.

    For r=8,9,10r = 8, 9, 10 the left-handed factor (110)10−r\left(\frac{1}{10}\right)^{10-r} is 1100\frac{1}{100}⁠, 110\frac{1}{10} and 1, and each numerator is 10Cr×9r{}^{10}C_r \times 9^r over the same 101010^{10}⁠.
    10C8=45{}^{10}C_8 = 45⁠, 10C9=10{}^{10}C_9 = 10⁠, 10C10=1{}^{10}C_{10} = 1⁠. Basics: counting

  6. Add, then subtract from 1

    P(X≥7)=∑r=71010Cr(0.9)r(0.1)10−r=573956280+19371024451010+3874204890+34867844011010=98720480161010.\begin{aligned} &P(X \ge 7) \\ &= \sum_{r=7}^{10} {}^{10}C_r (0.9)^r (0.1)^{10-r} \\ &= \frac{573956280 + 1937102445}{10^{10}} \\ &\quad + \frac{3874204890 + 3486784401}{10^{10}} \\ &= \frac{9872048016}{10^{10}}. \end{aligned} P(X≤6)=1−98720480161010=1279519841010=7996999625000000≈0.0128.\begin{aligned} P(X \le 6) &= 1 - \frac{9872048016}{10^{10}} \\ &= \frac{127951984}{10^{10}} \\ &= \frac{7996999}{625000000} \\ &\approx 0.0128. \end{aligned}
    The four cases are mutually exclusive, so they add. In the last step both 127951984 and 101010^{10} are divided by 16, their highest common factor; 79969997996999 is odd and does not end in 0 or 5, so the fraction is in lowest terms.
    P(E′)=1−P(E)P(E') = 1 - P(E)⁠. Basics: rules of probability

  7. A sense check

    With 9 in 10 people right-handed, a sample of 10 usually has 9 or 10 right-handers (together about 0.3874+0.3487=0.73610.3874 + 0.3487 = 0.7361⁠). Having 6 or fewer is far below that, so a small answer, about 1.3%, makes sense. Adding the seven cases r=0r = 0 to 66 directly gives the same 0.01279519840.0127951984⁠; the largest of them is P(X=6)=210×961010≈0.0112P(X = 6) = \frac{210 \times 9^6}{10^{10}} \approx 0.0112⁠.

    Two routes to the same number, the complement and the direct sum, confirm the arithmetic.
    The probabilities of all the cases r=0r = 0 to 1010 add to 1. Basics: rules of probability
Result

P(at most 6 of the 10 are right-handed) =1−∑r=71010Cr(0.9)r(0.1)10−r= 1 - \sum_{r=7}^{10} {}^{10}C_r (0.9)^r (0.1)^{10-r} =7996999625000000≈0.0128= \frac{7996999}{625000000} \approx 0.0128⁠.

(Textbook answer: 1−∑r=71010Cr(0.9)r(0.1)10−r1 - \sum_{r=7}^{10} {}^{10}C_r (0.9)^r (0.1)^{10-r}⁠.)

This question uses the binomial distribution, which is not in the current CBSE syllabus.

Check it with numbers

Add the four decimals from Step 5: 0.0574+0.1937=0.25110.0574 + 0.1937 = 0.2511⁠, 0.3874+0.3487=0.73610.3874 + 0.3487 = 0.7361 and 0.2511+0.7361=0.98720.2511 + 0.7361 = 0.9872⁠; then 1−0.9872=0.01281 - 0.9872 = 0.0128⁠, the same as 7996999625000000=0.0127951984\frac{7996999}{625000000} = 0.0127951984 to four places. The verify script also lists all 210=10242^{10} = 1024 patterns of R and L, gives each its product probability, and finds that the patterns with at most 6 R's have total probability exactly 7996999625000000\frac{7996999}{625000000}⁠.

Common mistakes
  1. Finding only "exactly 6". "At most 6" also includes 0, 1, 2, 3, 4 and 5.
  2. Using 1−P(X≥6)1 - P(X \ge 6)⁠. "At most 6" includes 6, so its complement starts at 7.
  3. Leaving out 10Cr{}^{10}C_r⁠. (0.9)7(0.1)3(0.9)^7(0.1)^3 is the probability of one pattern; there are 120 patterns with 7 right-handers.
  4. Swapping the powers. The power of 0.9 counts the right-handers, the power of 0.1 the others; the two powers add to 10.
What to take away
  1. For repeated independent trials: probability of one pattern = product; number of patterns with rr successes = nCr{}^{n}C_r⁠; so P(exactly rr⁠) =nCr pr(1−p)n−r= {}^{n}C_r\, p^r (1-p)^{n-r}⁠.
  2. "At most" and "at least" questions are often quicker by the complement.
  3. The same multiplication of independent trials appears in book Example 23, where two players throw a die in turn.

Every fact used, and where it comes from

FactUsed inSource
k%=k100k\% = \frac{k}{100}⁠; P(E′)=1−P(E)P(E') = 1 - P(E)Steps 1, 2, 6Class XI (Basics: rules of probability)
Independent experiments: probabilities multiplySteps 1, 3Section 13.4, Remarks (iii) and (iv) (Lesson: independent events)
nCr=n!r! (n−r)!{}^{n}C_r = \frac{n!}{r!\,(n-r)!}Steps 4, 5, 7Class XI, Permutations and Combinations (Basics: counting)
Mutually exclusive events addSteps 4, 6Class XI axioms (Basics: rules of probability)
P(exactly rr⁠) =nCr pr(1−p)n−r= {}^{n}C_r\, p^r (1-p)^{n-r} (the binomial distribution)Steps 4, 5Built here from the rules above; not taught in this reprint
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →