SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Probability

38 questions · 38 still being checked

EXERCISE 14.2 11–21 (part 3 of 4)

  1. Exercise 11

    In a lottery, a person choses six different natural numbers at random from 1\displaystyle 1 to 20\displaystyle 20 , and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? [Hint order of the numbers is not important.]

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    NCERT’s answer
    $\displaystyle \frac{1}{38760}$
    Combinations, since order does not matter. The person picks six different numbers from \(\displaystyle \{1, 2, \dots, 20\} \) and the order in which they are chosen is irrelevant, so the number of possible selections is\[n(S) = \binom{20}{6} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \]\[= \frac{27907200}{720} = 38760 \]Exactly one of these $\displaystyle 38760$ selections matches the six numbers fixed by the committee, so the winning event has \(\displaystyle n(E) = 1 \), and all selections are equally likely:\[P(\text{winning}) = \frac{n(E)}{n(S)} = \frac{1}{38760} \]The probability of winning the prize is \(\displaystyle \dfrac{1}{\binom{20}{6}} = \dfrac{1}{38760} \).

  2. Exercise 12

    Check whether the following probabilities P(A) and P(B) are consistently defined
    (i)
    P(A)=0.5,P(B)=0.7,P(A∩B)=0.6\displaystyle \mathrm{P}(\mathrm{A})=0.5, \mathrm{P}(\mathrm{B})=0.7, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.6
    (ii)
    P(A)=0.5,P(B)=0.4,P(A∪B)=0.8\displaystyle \mathrm{P}(\mathrm{A})=0.5, \mathrm{P}(\mathrm{B})=0.4, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.8

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    NCERT’s answer
    (i)
    No, because $\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})$ must be less than or equal to $\displaystyle \mathrm{P}(\mathrm{A})$ and $\displaystyle \mathrm{P}(\mathrm{B})$, (ii) Yes
    Consistency check. A pair of probabilities is consistently defined only if it can actually arise from some sample space. Two tests settle it: since \(\displaystyle A\cap B\subseteq A\) and \(\displaystyle A\cap B\subseteq B\), \[P(A\cap B)\le \min\{P(A),\,P(B)\},\] and the addition rule \(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\) must return a value with \(\displaystyle \max\{P(A),P(B)\}\le P(A\cup B)\le 1\).(i) \(\displaystyle P(A)=0.5,\ P(B)=0.7,\ P(A\cap B)=0.6\).Here \(\displaystyle P(A\cap B)=0.6\) while \(\displaystyle P(A)=0.5\), so \[P(A\cap B) > P(A),\] which is impossible for a subset of \(\displaystyle A\). Not consistently defined.(ii) \(\displaystyle P(A)=0.5,\ P(B)=0.4,\ P(A\cup B)=0.8\).The addition rule forces \[P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.5+0.4-0.8=0.1 .\] This is non-negative and satisfies \(\displaystyle 0.1\le 0.5\) and \(\displaystyle 0.1\le 0.4\); also \(\displaystyle 0.8\le 1\) and \(\displaystyle 0.8\ge 0.5=\max\{P(A),P(B)\}\). Every requirement holds. Consistently defined.

  3. Exercise 13

    Fill in the blanks in following table:
    P(A)P(B)P(A∩B)\displaystyle \mathbf{P}(\mathbf{A} \cap \mathbf{B})P(A∪B)\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})
    (i)13\displaystyle \frac{1}{3}15\displaystyle \frac{1}{5}115\displaystyle \frac{1}{15}⋯
    (ii)0.35\displaystyle 0.35...0.25\displaystyle 0.250.6\displaystyle 0.6
    (iii)0.5\displaystyle 0.50.35\displaystyle 0.35⋯0.7\displaystyle 0.7

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    NCERT’s answer
    (i)
    $\displaystyle \frac{7}{15}$ (ii) $\displaystyle 0.5$ (iii) $\displaystyle 0.15$
    Addition rule. For any two events, \[P(A\cup B)=P(A)+P(B)-P(A\cap B).\] Each blank is found by rearranging this one identity.(i) \(\displaystyle P(A)=\dfrac13,\ P(B)=\dfrac15,\ P(A\cap B)=\dfrac1{15}\): \[P(A\cup B)=\frac13+\frac15-\frac1{15}=\frac{5+3-1}{15}=\frac{7}{15}.\](ii) \(\displaystyle P(A)=0.35,\ P(A\cap B)=0.25,\ P(A\cup B)=0.6\). Rearranging, \[P(B)=P(A\cup B)+P(A\cap B)-P(A)=0.6+0.25-0.35=0.5 .\](iii) \(\displaystyle P(A)=0.5,\ P(B)=0.35,\ P(A\cup B)=0.7\). Rearranging, \[P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.5+0.35-0.7=0.15 .\]Answers: (i) \(\displaystyle P(A\cup B)=\dfrac{7}{15}\) (ii) \(\displaystyle P(B)=0.5\) (iii) \(\displaystyle P(A\cap B)=0.15\).

  4. Exercise 14

    Given P(A)=35\displaystyle \mathrm{P}(\mathrm{A})=\frac{3}{5} and P(B)=15\displaystyle \mathrm{P}(\mathrm{B})=\frac{1}{5}. Find P(A or B), if A and B are mutually exclusive events.

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    Addition rule for mutually exclusive events. If \(\displaystyle A\) and \(\displaystyle B\) are mutually exclusive then \(\displaystyle A\cap B=\varnothing\), so \(\displaystyle P(A\cap B)=0\) and the addition rule collapses to \[P(A\cup B)=P(A)+P(B).\]Substituting \(\displaystyle P(A)=\dfrac35\) and \(\displaystyle P(B)=\dfrac15\), \[P(A\text{ or }B)=\frac35+\frac15=\frac45 .\]\(\displaystyle P(A\text{ or }B)=\dfrac{4}{5}\).

  5. Exercise 15

    If E and F are events such that P(E)=14,P(F)=12\displaystyle \mathrm{P}(\mathrm{E})=\frac{1}{4}, \mathrm{P}(\mathrm{F})=\frac{1}{2} and P(E\displaystyle \mathrm{P}(\mathrm{E} and F)=18\displaystyle )=\frac{1}{8}, find
    (i)
    P(E or F),
    (ii)
    P(not E and not F).

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    NCERT’s answer
    (i)
    $\displaystyle \frac{5}{8}$ (ii) $\displaystyle \frac{3}{8}$
    Addition rule and De Morgan's law. Write \(\displaystyle P(E)=\dfrac14,\ P(F)=\dfrac12,\ P(E\cap F)=\dfrac18\).(i) \(\displaystyle P(E\text{ or }F)=P(E\cup F)=P(E)+P(F)-P(E\cap F)\): \[P(E\cup F)=\frac14+\frac12-\frac18=\frac{2+4-1}{8}=\frac58 .\](ii) "Not \(\displaystyle E\) and not \(\displaystyle F\)" is \(\displaystyle E'\cap F'\), which by De Morgan's law equals \(\displaystyle (E\cup F)'\). Hence \[P(E'\cap F')=1-P(E\cup F)=1-\frac58=\frac38 .\]Answers: (i) \(\displaystyle P(E\text{ or }F)=\dfrac58\) (ii) \(\displaystyle P(\text{not }E\text{ and not }F)=\dfrac38\).

  6. Exercise 16

    Events E and F are such that P (not E or not F)=0.25\displaystyle )=0.25, State whether E and F are mutually exclusive.

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    NCERT’s answer
    No
    De Morgan's law. "Not \(\displaystyle E\) or not \(\displaystyle F\)" is the event \(\displaystyle E'\cup F'\), and De Morgan's law gives \[E'\cup F'=(E\cap F)'.\]Therefore \[P(E\cap F)=1-P\big((E\cap F)'\big)=1-P(E'\cup F')=1-0.25=0.75 .\]Events are mutually exclusive precisely when \(\displaystyle P(E\cap F)=0\). Here \(\displaystyle P(E\cap F)=0.75\neq 0\), so \(\displaystyle E\) and \(\displaystyle F\) can occur together.No — \(\displaystyle E\) and \(\displaystyle F\) are not mutually exclusive, since \(\displaystyle P(E\cap F)=0.75\neq 0\).

  7. Exercise 17

    A and B are events such that P(A)=0.42,P(B)=0.48\displaystyle \mathrm{P}(\mathrm{A})=0.42, \mathrm{P}(\mathrm{B})=0.48 and P(A\displaystyle \mathrm{P}(\mathrm{A} and B)=0.16\displaystyle )=0.16. Determine
    (i)
    P(not A),
    (ii)
    P(not B) and
    (iii)
    P(A or B)

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    NCERT’s answer
    (i)
    0.$\displaystyle 58$ (ii) $\displaystyle 0.52$ (iii) $\displaystyle 0.74$
    Complement rule and addition rule. Given \(\displaystyle P(A)=0.42,\ P(B)=0.48,\ P(A\cap B)=0.16\).(i) The complement rule \(\displaystyle P(A')=1-P(A)\) gives \[P(\text{not }A)=1-0.42=0.58 .\](ii) Likewise \[P(\text{not }B)=1-0.48=0.52 .\](iii) The addition rule gives \[P(A\text{ or }B)=P(A)+P(B)-P(A\cap B)=0.42+0.48-0.16=0.74 .\]Answers: (i) \(\displaystyle 0.58\) (ii) \(\displaystyle 0.52\) (iii) \(\displaystyle 0.74\).

  8. Exercise 18

    In Class XI of a school 40\displaystyle 40% of the students study Mathematics and 30\displaystyle 30% study Biology. 10\displaystyle 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

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    NCERT’s answer
    0.$\displaystyle 6$
    Addition rule. Let \(\displaystyle M\) be the event that the chosen student studies Mathematics and \(\displaystyle B\) that the student studies Biology. Reading the percentages as probabilities for a randomly chosen student, \[P(M)=0.40,\qquad P(B)=0.30,\qquad P(M\cap B)=0.10 .\]The $\displaystyle 10$% who study both are counted in each of the first two figures, so they must be subtracted once: \[P(M\cup B)=P(M)+P(B)-P(M\cap B)=0.40+0.30-0.10=0.60 .\]The probability that the student studies Mathematics or Biology is \(\displaystyle 0.6\).

  9. Exercise 19

    In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8\displaystyle 0.8 and the probability of passing the second examination is 0.7\displaystyle 0.7 . The probability of passing atleast one of them is 0.95. What is the probability of passing both?

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    NCERT’s answer
    0.$\displaystyle 55$
    Addition rule, rearranged. Let \(\displaystyle A\) be the event of passing the first examination and \(\displaystyle B\) that of passing the second. Then \[P(A)=0.8,\qquad P(B)=0.7,\qquad P(A\cup B)=0.95,\] since "passing at least one" is exactly \(\displaystyle A\cup B\).From \(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\), \[P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.8+0.7-0.95=0.55 .\]The probability of passing both examinations is \(\displaystyle 0.55\).

  10. Exercise 20

    The probability that a student will pass the final examination in both English and Hindi is 0.5\displaystyle 0.5 and the probability of passing neither is 0.1\displaystyle 0.1 . If the probability of passing the English examination is 0.75\displaystyle 0.75 , what is the probability of passing the Hindi examination?

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    NCERT’s answer
    0.$\displaystyle 65$
    Addition rule with a complement. Let \(\displaystyle E\) be the event of passing English and \(\displaystyle H\) that of passing Hindi. We are told \[P(E\cap H)=0.5,\qquad P(E'\cap H')=0.1,\qquad P(E)=0.75 .\]"Passing neither" is \(\displaystyle E'\cap H'=(E\cup H)'\) by De Morgan's law, so \[P(E\cup H)=1-0.1=0.9 .\]Now apply \(\displaystyle P(E\cup H)=P(E)+P(H)-P(E\cap H)\): \[0.9=0.75+P(H)-0.5\ \Longrightarrow\ P(H)=0.9-0.25=0.65 .\]The probability of passing the Hindi examination is \(\displaystyle 0.65\).

  11. Exercise 21

    In a class of 60\displaystyle 60 students, 30\displaystyle 30 opted for NCC, 32\displaystyle 32 opted for NSS and 24\displaystyle 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that
    (i)
    The student opted for NCC or NSS.
    (ii)
    The student has opted neither NCC nor NSS.
    (iii)
    The student has opted NSS but not NCC.

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    NCERT’s answer
    (i)
    $\displaystyle \frac{19}{30}$ (ii) $\displaystyle \frac{11}{30}$ (iii) $\displaystyle \frac{2}{15}$
    Addition rule on a class of 60. Let \(\displaystyle C\) be the event that the selected student opted for NCC and \(\displaystyle S\) that the student opted for NSS. With equally likely selection from $\displaystyle 60$ students, \[P(C)=\frac{30}{60},\qquad P(S)=\frac{32}{60},\qquad P(C\cap S)=\frac{24}{60}.\](i) NCC or NSS. \[P(C\cup S)=\frac{30}{60}+\frac{32}{60}-\frac{24}{60}=\frac{38}{60}=\frac{19}{30}.\](ii) Neither NCC nor NSS. This is \(\displaystyle (C\cup S)'\), so \[P(C'\cap S')=1-\frac{19}{30}=\frac{11}{30}.\](iii) NSS but not NCC. The students who took NSS split into those who also took NCC and those who did not, so \(\displaystyle P(S\cap C')=P(S)-P(S\cap C)\): \[P(S\cap C')=\frac{32}{60}-\frac{24}{60}=\frac{8}{60}=\frac{2}{15}.\] (Equivalently, \(\displaystyle 32-24=8\) students took NSS only.) A Venn diagram of the two overlapping groups — $\displaystyle 6$ NCC only, $\displaystyle 24$ both, $\displaystyle 8$ NSS only, $\displaystyle 22$ neither — shows all three answers at a glance.Answers: (i) \(\displaystyle \dfrac{19}{30}\) (ii) \(\displaystyle \dfrac{11}{30}\) (iii) \(\displaystyle \dfrac{2}{15}\).