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NCERT Exemplar · Class 12 Mathematics Probability

108 questions · 108 still being checked

EXERCISE 13.3 31–40 (part 4 of 11)

  1. Exercise 31

    A factory produces bulbs. The probability that any one bulb is defective is 150\displaystyle \frac{1}{50} and they are packed in boxes of 10. From a single box, find the probability that
    (i)
    none of the bulbs is defective
    (ii)
    exactly two bulbs are defective
    (iii)
    more than 8\displaystyle 8 bulbs work properly

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    NCERT’s answer
    (i)
    \(\displaystyle \left(\frac{49}{50}\right)^{10}\)
    (ii)
    \(\displaystyle \frac{45(49)^8}{(50)^{10}}\)
    (iii)
    \(\displaystyle \frac{59(49)^9}{(50)^{10}}\)
    Let \(\displaystyle X\) be the number of defective bulbs in a box.
    \[X \sim B\left(10, \tfrac{1}{50}\right), \qquad p = \frac{1}{50},\; q = \frac{49}{50} \]
    (i)
    \[P(X=0) = \left(\frac{49}{50}\right)^{10} \]
    (ii)
    \[P(X=2) = \binom{10}{2}\left(\frac{1}{50}\right)^2\left(\frac{49}{50}\right)^8 = \frac{45\cdot 49^8}{50^{10}} \]
    (iii)
    More than $\displaystyle 8$ working means at most $\displaystyle 1$ defective.
    \[P(X \le 1) = \left(\frac{49}{50}\right)^{10} + 10\cdot\frac{1}{50}\left(\frac{49}{50}\right)^9 \]
    \[= \left(\frac{49}{50}\right)^9\left(\frac{49}{50} + \frac{10}{50}\right) = \frac{59}{50}\left(\frac{49}{50}\right)^9 \]
    Answer: (i) \(\displaystyle \left(\frac{49}{50}\right)^{10} \approx 0.817\); (ii) \(\displaystyle \frac{45\cdot 49^8}{50^{10}} \approx 0.0153\); (iii) \(\displaystyle \frac{59}{50}\left(\frac{49}{50}\right)^9 \approx 0.984\).

  2. Exercise 32

    Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2\displaystyle 2-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?

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    NCERT’s answer
    \(\displaystyle \frac{1}{3}\)
    Let \(\displaystyle F\) = fair coin, \(\displaystyle T\) = two-headed coin, \(\displaystyle H\) = head. By Bayes' theorem: \[P(F) = P(T) = \frac12, \qquad P(H \mid F) = \frac12, \qquad P(H \mid T) = 1 \] \[P(F \mid H) = \frac{P(F)\,P(H \mid F)}{P(F)\,P(H \mid F) + P(T)\,P(H \mid T)} \] \[= \frac{\frac12\cdot\frac12}{\frac12\cdot\frac12 + \frac12\cdot 1} = \frac{1/4}{3/4} = \frac13 \] Answer: \(\displaystyle \frac13\).

  3. Exercise 33

    Suppose that 6%\displaystyle 6 \% of the people with blood group O are left handed and 10%\displaystyle 10 \% of those with other blood groups are left handed 30%\displaystyle 30 \% of the people have blood group O. If a left handed person is selected at random, what is the probability that he/she will have blood group O?

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    NCERT’s answer
    \(\displaystyle \frac{9}{44}\)
    Let \(\displaystyle O\): blood group O, \(\displaystyle L\): left handed. \[P(O)=0.3,\quad P(O')=0.7 \] \[P(L\mid O)=0.06,\quad P(L\mid O')=0.10 \] \[P(O\mid L)=\frac{P(O)\,P(L\mid O)}{P(O)\,P(L\mid O)+P(O')\,P(L\mid O')} \quad \text{(Bayes' theorem)} \] \[=\frac{0.3\times0.06}{0.3\times0.06+0.7\times0.10} \] \[=\frac{0.018}{0.018+0.070}=\frac{0.018}{0.088}=\frac{9}{44} \] Answer: \(\displaystyle \dfrac{9}{44}\)

  4. Exercise 34

    Two natural numbers r,s\displaystyle r, s are drawn one at a time, without replacement from the set S={1,2,3,…,n}\displaystyle \mathrm{S}=\{1,2,3, \ldots, n\}. Find P[r≤p∣s≤p]\displaystyle \mathrm{P}[r \leq p \mid s \leq p], where p∈S\displaystyle p \in S.

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    NCERT’s answer
    \(\displaystyle \frac{p-1}{n-1}\)
    Let \(\displaystyle E: r\le p\) and \(\displaystyle F: s\le p\); the draws are without replacement. \[P(F)=\frac{p}{n} \] \[P(E\cap F)=\frac{p}{n}\cdot\frac{p-1}{n-1} \quad \text{(both from the } p \text{ numbers } \le p\text{, no repeat)} \] \[P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{p(p-1)}{n(n-1)}\cdot\frac{n}{p} \] \[=\frac{p-1}{n-1} \] Answer: \(\displaystyle \dfrac{p-1}{n-1}\)

  5. Exercise 35

    Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.

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    NCERT’s answer
    X$\displaystyle 1$$\displaystyle 2$$\displaystyle 3$$\displaystyle 4$$\displaystyle 5$$\displaystyle 6$
    P(X)$\displaystyle 36$$\displaystyle 36$$\displaystyle 36$$\displaystyle 36$$\displaystyle 36$$\displaystyle 36$
    Let \(\displaystyle X\) be the larger score; the $\displaystyle 36$ pairs are equally likely. \[P(X\le k)=\frac{k^2}{36} \quad \text{(both scores} \le k\text{)} \] \[P(X=k)=\frac{k^2-(k-1)^2}{36}=\frac{2k-1}{36} \] \[\begin{array}{c|cccccc} X & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(X) & \frac{1}{36} & \frac{3}{36} & \frac{5}{36} & \frac{7}{36} & \frac{9}{36} & \frac{11}{36} \end{array} \] \[E(X)=\sum x\,P(X=x)=\frac{1\cdot1+2\cdot3+3\cdot5+4\cdot7+5\cdot9+6\cdot11}{36} \] \[=\frac{1+6+15+28+45+66}{36}=\frac{161}{36} \] Answer: distribution as in the table; mean \(\displaystyle =\dfrac{161}{36}\)

  6. Exercise 36

    The random variable X can take only the values 0,1,2\displaystyle 0,1,2. Given that P(X=0)=\displaystyle \mathrm{P}(\mathrm{X}=0)= P(X=1)=p\displaystyle \mathrm{P}(\mathrm{X}=1)=p and that E(X2)=E[X]\displaystyle \mathrm{E}\left(\mathrm{X}^2\right)=\mathrm{E}[\mathrm{X}], find the value of p\displaystyle p.

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    NCERT’s answer
    \(\displaystyle p=\frac{1}{2}\)
    The probabilities sum to $\displaystyle 1$, so \(\displaystyle P(X=2)=1-2p\). \[E(X)=0\cdot p+1\cdot p+2(1-2p)=2-3p \] \[E(X^2)=0\cdot p+1\cdot p+4(1-2p)=4-7p \] \[E(X^2)=E(X)\ \Rightarrow\ 4-7p=2-3p \] \[4p=2 \] \[p=\frac12 \] Answer: \(\displaystyle p=\dfrac12\)

  7. Exercise 37

    Find the variance of the distribution:
    x\displaystyle x0\displaystyle 01\displaystyle 12\displaystyle 23\displaystyle 34\displaystyle 45\displaystyle 5
    P(x)\displaystyle \mathrm{P}(x)16\displaystyle \frac{1}{6}518\displaystyle \frac{5}{18}29\displaystyle \frac{2}{9}16\displaystyle \frac{1}{6}19\displaystyle \frac{1}{9}118\displaystyle \frac{1}{18}

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    NCERT’s answer
    \(\displaystyle \frac{665}{324}\)
    Write each probability over 18. \[P(x)=\frac{3}{18},\ \frac{5}{18},\ \frac{4}{18},\ \frac{3}{18},\ \frac{2}{18},\ \frac{1}{18} \] \[E(X)=\sum xP(x)=\frac{0\cdot3+1\cdot5+2\cdot4+3\cdot3+4\cdot2+5\cdot1}{18}=\frac{35}{18} \] \[E(X^2)=\sum x^2P(x)=\frac{0+1\cdot5+4\cdot4+9\cdot3+16\cdot2+25\cdot1}{18}=\frac{105}{18}=\frac{35}{6} \] \[\operatorname{Var}(X)=E(X^2)-[E(X)]^2 \] \[=\frac{35}{6}-\frac{1225}{324}=\frac{1890-1225}{324}=\frac{665}{324} \] Answer: \(\displaystyle \operatorname{Var}(X)=\dfrac{665}{324}\)

  8. Exercise 38

    A and B throw a pair of dice alternately. A wins the game if he gets a total of 6\displaystyle 6 and B wins if she gets a total of 7. It A starts the game, find the probability of winning the game by A in third throw of the pair of dice.

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    NCERT’s answer
    \(\displaystyle \frac{775}{7776}\)
    A wins on the third throw only if A fails, B fails, then A scores 6. \[P(\text{total }6)=\frac{5}{36},\qquad P(\text{total }7)=\frac{6}{36}=\frac16 \] \[P=\left(1-\frac{5}{36}\right)\left(1-\frac16\right)\left(\frac{5}{36}\right) \] \[=\frac{31}{36}\cdot\frac{5}{6}\cdot\frac{5}{36}=\frac{775}{7776} \] Answer: \(\displaystyle \dfrac{775}{7776}\)

  9. Exercise 39

    Two dice are tossed. Find whether the following two events A and B are independent: A={(x,y):x+y=11}B={(x,y):x≠5}\mathrm{A}=\{(x, y): x+y=11\} \quad \mathrm{B}=\{(x, y): x \neq 5\} where (x,y)\displaystyle (x, y) denotes a typical sample point.

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    NCERT’s answer
    not independent
    \[A=\{(5,6),(6,5)\} \ \Rightarrow\ P(A)=\frac{2}{36}=\frac1{18} \] \[B:\ x\ne5 \ \Rightarrow\ P(B)=\frac{30}{36}=\frac56 \] \[A\cap B=\{(6,5)\} \ \Rightarrow\ P(A\cap B)=\frac1{36} \] \[P(A)\,P(B)=\frac1{18}\cdot\frac56=\frac{5}{108} \] \[P(A\cap B)=\frac{3}{108}\ne P(A)\,P(B) \] Answer: \(\displaystyle A\) and \(\displaystyle B\) are not independent.

  10. Exercise 40

    An urn contains m\displaystyle m white and n\displaystyle n black balls. A ball is drawn at random and is put back into the urn along with k\displaystyle k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on k\displaystyle k.

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    Let \(\displaystyle W_1, B_1\): first ball white, black; \(\displaystyle W_2\): second ball white. \[P(W_1)=\frac{m}{m+n},\qquad P(B_1)=\frac{n}{m+n} \] \[P(W_2\mid W_1)=\frac{m+k}{m+n+k},\qquad P(W_2\mid B_1)=\frac{m}{m+n+k} \] \[P(W_2)=P(W_1)P(W_2\mid W_1)+P(B_1)P(W_2\mid B_1) \quad \text{(total probability)} \] \[=\frac{m(m+k)+nm}{(m+n)(m+n+k)}=\frac{m(m+n+k)}{(m+n)(m+n+k)} \] \[=\frac{m}{m+n} \] Answer: \(\displaystyle P(W_2)=\dfrac{m}{m+n}\), which does not involve \(\displaystyle k\).