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NCERT Solutions · Class 12 Mathematics Probability

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EXERCISE 13.1 1–10 (part 1 of 7)

  1. Exercise 1

    Given that E and F are events such that \(\displaystyle \mathrm{P}(\mathrm{E})=0.6, \mathrm{P}(\mathrm{F})=0.3\) and \(\displaystyle \mathrm{P}(\mathrm{E} \cap \mathrm{F})=0.2\), find \(\displaystyle \mathrm{P}(\mathrm{E} \mid \mathrm{F})\) and \(\displaystyle \mathrm{P}(\mathrm{F} \mid \mathrm{E})\)

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    NCERT’s answer
    \(\displaystyle \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{2}{3}, \mathrm{P}(\mathrm{F} \mid \mathrm{E})=\frac{1}{3}\)
    By the definition of conditional probability, for \(\displaystyle P(F)\neq 0\), \[P(E\mid F)=\frac{P(E\cap F)}{P(F)}\] Here \(\displaystyle P(E)=0.6,\ P(F)=0.3,\ P(E\cap F)=0.2\), and both \(\displaystyle P(E)\) and \(\displaystyle P(F)\) are non-zero, so both conditional probabilities are defined. \[P(E\mid F)=\frac{0.2}{0.3}=\frac{2}{3}\] \[P(F\mid E)=\frac{P(E\cap F)}{P(E)}=\frac{0.2}{0.6}=\frac{1}{3}\] Hence \(\displaystyle P(E\mid F)=\frac{2}{3}\) and \(\displaystyle P(F\mid E)=\frac{1}{3}\).

  2. Exercise 2

    Compute \(\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})\), if \(\displaystyle \mathrm{P}(\mathrm{B})=0.5\) and \(\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.32\)

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    NCERT’s answer
    \(\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{16}{25}\)
    Using the definition of conditional probability, valid since \(\displaystyle P(B)=0.5\neq 0\), \[P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.32}{0.5}=0.64\] Hence \(\displaystyle P(A\mid B)=0.64=\frac{16}{25}\).

  3. Exercise 3

    If \(\displaystyle \mathrm{P}(\mathrm{A})=0.8, \mathrm{P}(\mathrm{B})=0.5\) and \(\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A})=0.4\), find
    (i)
    \(\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})\)
    (ii)
    \(\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})\)
    (iii)
    \(\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})\)

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    NCERT’s answer
    (i)
    0.$\displaystyle 32$ (ii) $\displaystyle 0.64$ (iii) $\displaystyle 0.98$
    Given \(\displaystyle P(A)=0.8,\ P(B)=0.5,\ P(B\mid A)=0.4\).
    (i)
    From \(\displaystyle P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}\), multiply across by \(\displaystyle P(A)\) (the multiplication rule):
    \[P(A\cap B)=P(B\mid A)\,P(A)=0.4\times 0.8=0.32\]
    (ii)
    Since \(\displaystyle P(B)=0.5\neq 0\),
    \[P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.32}{0.5}=0.64\]
    (iii)
    By the addition theorem \(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\),
    \[P(A\cup B)=0.8+0.5-0.32=0.98\]
    Hence \(\displaystyle P(A\cap B)=0.32\), \(\displaystyle P(A\mid B)=0.64\), \(\displaystyle P(A\cup B)=0.98\).

  4. Exercise 4

    Evaluate \(\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})\), if \(\displaystyle 2 \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})=\frac{5}{13}\) and \(\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{2}{5}\)

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    NCERT’s answer
    \(\displaystyle \frac{11}{26}\)
    From \(\displaystyle 2P(A)=P(B)=\frac{5}{13}\): \[P(B)=\frac{5}{13},\qquad P(A)=\frac{1}{2}\cdot\frac{5}{13}=\frac{5}{26}\] The conditional probability \(\displaystyle P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\) gives, on multiplying by \(\displaystyle P(B)\), \[P(A\cap B)=P(A\mid B)\,P(B)=\frac{2}{5}\times\frac{5}{13}=\frac{2}{13}\] By the addition theorem, \[P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}=\frac{5}{26}+\frac{10}{26}-\frac{4}{26}=\frac{11}{26}\] Hence \(\displaystyle P(A\cup B)=\dfrac{11}{26}\).

  5. Exercise 5

    If \(\displaystyle \mathrm{P}(\mathrm{A})=\frac{6}{11}, \mathrm{P}(\mathrm{B})=\frac{5}{11}\) and \(\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}\), find
    (i)
    \(\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})\)
    (ii)
    \(\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})\)
    (iii)
    \(\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A})\)

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{4}{11}\) (ii) \(\displaystyle \frac{4}{5}\) (iii) \(\displaystyle \frac{2}{3}\)
    Given \(\displaystyle P(A)=\frac{6}{11},\ P(B)=\frac{5}{11},\ P(A\cup B)=\frac{7}{11}\).
    (i)
    Rearranging the addition theorem \(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\),
    \[P(A\cap B)=P(A)+P(B)-P(A\cup B)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{4}{11}\]
    (ii)
    Since \(\displaystyle P(B)\neq 0\),
    \[P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{4/11}{5/11}=\frac{4}{5}\]
    (iii)
    Since \(\displaystyle P(A)\neq 0\),
    \[P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{4/11}{6/11}=\frac{2}{3}\]
    Hence \(\displaystyle P(A\cap B)=\frac{4}{11}\), \(\displaystyle P(A\mid B)=\frac{4}{5}\), \(\displaystyle P(B\mid A)=\frac{2}{3}\).
  6. Determine P(EIF) in Exercises $\displaystyle 6$ to 9.

    Exercise 6

    A coin is tossed three times, where
    (i)
    E : head on third toss , F : heads on first two tosses
    (ii)
    E : at least two heads , F : at most two heads
    (iii)
    E : at most two tails , F : at least one tail

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{2}\) (ii) \(\displaystyle \frac{3}{7}\) (iii) \(\displaystyle \frac{6}{7}\)
    A coin tossed three times has \(\displaystyle 2^3=8\) equally likely outcomes:
    \[S=\{HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT\}\]
    Because the outcomes are equally likely,
    \[P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{n(E\cap F)/8}{n(F)/8}=\frac{n(E\cap F)}{n(F)}\]
    (i)
    \(\displaystyle E\) = head on the third toss \(\displaystyle =\{HHH,\ HTH,\ THH,\ TTH\}\); \(\displaystyle F\) = heads on the first two tosses \(\displaystyle =\{HHH,\ HHT\}\). The only outcome in both is \(\displaystyle HHH\), so \(\displaystyle E\cap F=\{HHH\}\).
    \[P(E\mid F)=\frac{1/8}{2/8}=\frac{1}{2}\]
    (ii)
    \(\displaystyle E\) = at least two heads \(\displaystyle =\{HHH,\ HHT,\ HTH,\ THH\}\); \(\displaystyle F\) = at most two heads = every outcome except \(\displaystyle HHH\), so \(\displaystyle n(F)=7\). Deleting \(\displaystyle HHH\) from \(\displaystyle E\) leaves \(\displaystyle E\cap F=\{HHT,\ HTH,\ THH\}\).
    \[P(E\mid F)=\frac{3/8}{7/8}=\frac{3}{7}\]
    (iii)
    \(\displaystyle E\) = at most two tails = every outcome except \(\displaystyle TTT\), so \(\displaystyle n(E)=7\); \(\displaystyle F\) = at least one tail = every outcome except \(\displaystyle HHH\), so \(\displaystyle n(F)=7\). An outcome lies in \(\displaystyle E\cap F\) exactly when it is neither \(\displaystyle TTT\) nor \(\displaystyle HHH\), so \(\displaystyle n(E\cap F)=6\).
    \[P(E\mid F)=\frac{6/8}{7/8}=\frac{6}{7}\]
    Hence the required values are \(\displaystyle \frac{1}{2}\), \(\displaystyle \frac{3}{7}\) and \(\displaystyle \frac{6}{7}\).

  7. Exercise 7

    Two coins are tossed once, where
    (i)
    E : tail appears on one coin, F : one coin shows head
    (ii)
    E : no tail appears, F : no head appears

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    NCERT’s answer
    (i)
    $\displaystyle 1$ (ii) $\displaystyle 0$
    Two coins tossed once give the equally likely sample space
    \[S=\{HH,\ HT,\ TH,\ TT\}\]
    (i)
    \(\displaystyle E\) = a tail appears on one coin \(\displaystyle =\{HT,\ TH\}\); \(\displaystyle F\) = one coin shows a head \(\displaystyle =\{HT,\ TH\}\). Then \(\displaystyle E\cap F=\{HT,\ TH\}\), so
    \[P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{2/4}{2/4}=1\]
    (Here \(\displaystyle E\) and \(\displaystyle F\) are the same event, so \(\displaystyle F\) occurring forces \(\displaystyle E\).)
    (ii)
    \(\displaystyle E\) = no tail appears \(\displaystyle =\{HH\}\); \(\displaystyle F\) = no head appears \(\displaystyle =\{TT\}\). These are mutually exclusive, so \(\displaystyle E\cap F=\phi\) and \(\displaystyle P(E\cap F)=0\). Since \(\displaystyle P(F)=\frac14\neq 0\),
    \[P(E\mid F)=\frac{0}{1/4}=0\]
    Hence \(\displaystyle P(E\mid F)=1\) in (i) and \(\displaystyle 0\) in (ii).

  8. Exercise 8

    A die is thrown three times, E : $\displaystyle 4$ appears on the third toss, F : $\displaystyle 6$ and $\displaystyle 5$ appears respectively on first two tosses

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    NCERT’s answer
    \(\displaystyle \frac{1}{6}\)
    A die thrown three times has \(\displaystyle 6^3=216\) equally likely outcomes, written as ordered triples \(\displaystyle (a,b,c)\).\(\displaystyle F\) = $\displaystyle 6$ on the first throw and $\displaystyle 5$ on the second \(\displaystyle =\{(6,5,c):c=1,2,\dots,6\}\), so \(\displaystyle n(F)=6\) and \(\displaystyle P(F)=\frac{6}{216}\).\(\displaystyle E\) = $\displaystyle 4$ on the third throw. The triples lying in both \(\displaystyle E\) and \(\displaystyle F\) must be \(\displaystyle (6,5,4)\), so \(\displaystyle n(E\cap F)=1\) and \(\displaystyle P(E\cap F)=\frac{1}{216}\). \[P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{1/216}{6/216}=\frac{1}{6}\] Hence \(\displaystyle P(E\mid F)=\dfrac{1}{6}\).

  9. Exercise 9

    Mother, father and son line up at random for a family picture E : son on one end, F : father in middle

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    NCERT’s answer
    $\displaystyle 1$
    Writing each arrangement left to right, the three people can line up in \(\displaystyle 3!=6\) equally likely ways: \[S=\{MFS,\ MSF,\ FMS,\ FSM,\ SMF,\ SFM\}\] where \(\displaystyle M\), \(\displaystyle F\), \(\displaystyle S\) denote mother, father, son.\(\displaystyle E\) = son on one end (first or third place) \(\displaystyle =\{MFS,\ FMS,\ SMF,\ SFM\}\), so \(\displaystyle P(E)=\frac{4}{6}\).\(\displaystyle F\) = father in the middle \(\displaystyle =\{MFS,\ SFM\}\), so \(\displaystyle P(F)=\frac{2}{6}\).In both arrangements of \(\displaystyle F\) the son is at an end, so \(\displaystyle E\cap F=\{MFS,\ SFM\}\) and \(\displaystyle P(E\cap F)=\frac{2}{6}\). \[P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{2/6}{2/6}=1\] Hence \(\displaystyle P(E\mid F)=1\): once the father is in the middle, the mother and son occupy the two ends, so the son is certainly at an end.

  10. Exercise 10

    A black and a red dice are rolled.
    (a)
    Find the conditional probability of obtaining a sum greater than $\displaystyle 9$ , given that the black die resulted in a 5.
    (b)
    Find the conditional probability of obtaining the sum $\displaystyle 8$, given that the red die resulted in a number less than 4.

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    NCERT’s answer
    (a)
    \(\displaystyle \frac{1}{3}\), (b) \(\displaystyle \frac{1}{9}\)
    Write an outcome as \(\displaystyle (b,r)\), \(\displaystyle b\) on the black die and \(\displaystyle r\) on the red die; there are \(\displaystyle 6\times 6=36\) equally likely outcomes.
    (a)
    Let \(\displaystyle A\) = 'sum greater than $\displaystyle 9$' and \(\displaystyle B\) = 'black die shows $\displaystyle 5$'.
    \(\displaystyle B=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}\), so \(\displaystyle P(B)=\frac{6}{36}\).
    From these, the sum exceeds $\displaystyle 9$ only for \(\displaystyle (5,5)\) (sum $\displaystyle 10$) and \(\displaystyle (5,6)\) (sum $\displaystyle 11$); \(\displaystyle (5,4)\) gives $\displaystyle 9$, which is not greater than 9. So \(\displaystyle P(A\cap B)=\frac{2}{36}\).
    \[P(A\mid B)=\frac{2/36}{6/36}=\frac{1}{3}\]
    (b)
    Let \(\displaystyle C\) = 'sum is $\displaystyle 8$' and \(\displaystyle D\) = 'red die shows a number less than $\displaystyle 4$', i.e. \(\displaystyle r\in\{1,2,3\}\). Then \(\displaystyle n(D)=6\times 3=18\), so \(\displaystyle P(D)=\frac{18}{36}\).
    For a sum of $\displaystyle 8$ with \(\displaystyle r<4\): \(\displaystyle r=2,b=6\) and \(\displaystyle r=3,b=5\); \(\displaystyle r=1\) would need \(\displaystyle b=7\), impossible. So \(\displaystyle P(C\cap D)=\frac{2}{36}\).
    \[P(C\mid D)=\frac{2/36}{18/36}=\frac{1}{9}\]
    Hence the answers are \(\displaystyle \frac{1}{3}\) and \(\displaystyle \frac{1}{9}\).