# NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.2) Exercise 13.2

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## NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.2) Exercise 13.2

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### Access NCERT Solutions for Class 12 Maths Chapter 13 – Probability

It is possible to solve all the questions of the Probability chapter with the help of  NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.2.

### NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.2

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Q.1 If P(A) = 3/5 and P (B) = 1/5, find P (A ∩ B) if A and B are independent events.

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right) \left[\begin{array}{l}\mathrm{As}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{independent}\\ \mathrm{events}\mathrm{.}\end{array}\right]\\ =\frac{3}{5}.\frac{1}{5}\\ =\frac{3}{25}\\ \mathrm{Thus},\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{3}{25}.\end{array}$

Q.2 Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Ans

$\begin{array}{l}\mathrm{Total}\mathrm{}\mathrm{number}\mathrm{}\mathrm{of}\mathrm{}\mathrm{cards}=52\\ \mathrm{Number}\mathrm{}\mathrm{of}\mathrm{}\mathrm{black}\mathrm{}\mathrm{cards}=26\\ \mathrm{Let}\mathrm{â€‹}\mathrm{A}=\mathrm{A}\mathrm{black}\mathrm{card}\mathrm{in}\mathrm{first}\mathrm{draw}\\ \mathrm{B}=\mathrm{A}\mathrm{black}\mathrm{card}\mathrm{in}\mathrm{second}\mathrm{draw}\\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{26}{52}=\frac{1}{2}\\ \mathrm{ }\mathrm{P}\left(\mathrm{B}\right)=\frac{25}{51}\\ \mathrm{Thus},\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{both}\mathrm{the}\mathrm{cards}\mathrm{black}\\ \mathrm{ }=\mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{ }=\frac{1}{2}×\frac{25}{51}\\ \mathrm{ }=\frac{25}{102}\end{array}$

Q.3 A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{A},\mathrm{B},\mathrm{and}\mathrm{C}\mathrm{be}\mathrm{the}\mathrm{respective}\mathrm{events}\mathrm{that}\mathrm{the}\mathrm{first},\mathrm{second},\\ \mathrm{and}\mathrm{third}\mathrm{drawn}\mathrm{orange}\mathrm{is}\mathrm{good}\mathrm{.}\\ \therefore \mathrm{ }\mathrm{Probability}\mathrm{that}\mathrm{first}\mathrm{drawn}\mathrm{orange}\mathrm{is}\mathrm{good},\mathrm{P}\left(\mathrm{A}\right)=\frac{12}{15}\\ \mathrm{Probability}\mathrm{that}\mathrm{second}\mathrm{drawn}\mathrm{orange}\mathrm{is}\mathrm{good},\mathrm{P}\left(\mathrm{B}\right)=\frac{11}{14}\\ ⇒\mathrm{Probability}\mathrm{that}\mathrm{third}\mathrm{drawn}\mathrm{orange}\mathrm{is}\mathrm{good},\mathrm{P}\left(\mathrm{C}\right)=\frac{10}{13}\\ \mathrm{The}\mathrm{box}\mathrm{is}\mathrm{approved}\mathrm{for}\mathrm{sale},\mathrm{if}\mathrm{all}\mathrm{the}\mathrm{three}\mathrm{oranges}\mathrm{are}\mathrm{good}\mathrm{.}\\ \mathrm{ }\mathrm{Thus},\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{all}\mathrm{the}\mathrm{oranges}\mathrm{good}=\mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right).\mathrm{P}\left(\mathrm{C}\right)\\ \mathrm{ }=\frac{12}{15}×\frac{11}{14}×\frac{10}{13}\\ \mathrm{ }=\frac{44}{91}\\ \mathrm{Therefore},\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{box}\mathrm{is}\mathrm{approved}\mathrm{for}\mathrm{sale}\mathrm{is} \frac{44}{91}.\end{array}$

Q.4 A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

Ans

$\begin{array}{l}\mathrm{Sample}\mathrm{space},\mathrm{when}\mathrm{a}\mathrm{fair}\mathrm{coin}\mathrm{and}\mathrm{an}\mathrm{unbiased}\mathrm{die}\mathrm{are}\mathrm{tossed},\\ \mathrm{ }\mathrm{S}=\left\{\begin{array}{l}\left(\mathrm{H},1\right),\left(\mathrm{H},2\right),\left(\mathrm{H},3\right),\left(\mathrm{H},4\right),\left(\mathrm{H},5\right),\left(\mathrm{H},6\right)\\ \left(\mathrm{T},1\right),\left(\mathrm{T},2\right),\left(\mathrm{T},3\right),\left(\mathrm{T},4\right),\left(\mathrm{T},5\right),\left(\mathrm{T},6\right)\end{array}\right\}\\ \mathrm{ }\mathrm{A}=\mathrm{Head}\mathrm{appears}\mathrm{on}\mathrm{the}\mathrm{coin}\\ =\left\{\left(\mathrm{H},1\right),\left(\mathrm{H},2\right),\left(\mathrm{H},3\right),\left(\mathrm{H},4\right),\left(\mathrm{H},5\right),\left(\mathrm{H},6\right)\right\}\\ \mathrm{ }\mathrm{B}=3\mathrm{on}\mathrm{the}\mathrm{die}\\ =\left\{\left(\mathrm{H},3\right),\left(\mathrm{T},3\right)\right\}\\ \mathrm{ }\mathrm{A}\cap \mathrm{B}=\left\{\left(\mathrm{H},3\right)\right\}\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{1}{12}\\ \mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)=\frac{6}{12}×\frac{2}{12}\\ =\frac{1}{12}\\ \mathrm{Since},\mathrm{ }\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{Therefore},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{independent}\mathrm{events}\mathrm{.}\end{array}$

Q.5 A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, the number is even, and B be the event, the number is red. Are A and B independent?

Ans

$\begin{array}{l}\mathrm{Sample}\mathrm{space},\mathrm{when}\mathrm{an}\mathrm{unbiased}\mathrm{die}\mathrm{are}\mathrm{tossed},\\ \mathrm{ }\mathrm{S}=\left\{1,2,3,4,5,6\right\}\\ \mathrm{ }\mathrm{A}=\mathrm{The}\mathrm{number}\mathrm{is}\mathrm{even}\\ =\left\{2, 4, 6\right\}\\ \mathrm{ }\mathrm{B}=\mathrm{The}\mathrm{number}\mathrm{is}\mathrm{red}\\ =\left\{4,5,6\right\}\\ \mathrm{ }\mathrm{A}\cap \mathrm{B}=\left\{4,6\right\}\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{2}{6}=\frac{1}{3}\\ \mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)=\frac{3}{6}×\frac{3}{6}\\ =\frac{1}{4}\\ \mathrm{Since},\mathrm{ }\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\ne \mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{Therefore},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{events}\mathrm{.}\end{array}$

Q.6 Let E and F be events with P (E) = 3/5, P(F) = 3/10 and P (E ∩ F) = 1/5 . Are E and F independent?

Ans

$\begin{array}{l}\mathrm{E}\mathrm{}\mathrm{and}\mathrm{}\mathrm{F}\mathrm{}\mathrm{be}\mathrm{}\mathrm{events}\mathrm{}\mathrm{with}\mathrm{}\mathrm{P}\left(\mathrm{E}\right)=\frac{3}{5},\mathrm{}\mathrm{P}\left(\mathrm{F}\right)=\frac{3}{10} \mathrm{and}\mathrm{}\mathrm{P}\mathrm{}\left(\mathrm{E}\cap \mathrm{F}\right)=1/5\\ \mathrm{P}\left(\mathrm{E}\right)×\mathrm{P}\left(\mathrm{F}\right)=\frac{3}{5}×\frac{3}{10}\\ =\frac{9}{50}\\ \mathrm{Since},\mathrm{ }\\ \mathrm{P}\mathrm{}\left(\mathrm{E}\cap \mathrm{F}\right)\ne \mathrm{P}\left(\mathrm{E}\right)×\mathrm{P}\left(\mathrm{F}\right)\\ \mathrm{Therefore},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{events}\mathrm{.}\end{array}$

Q.7 Given that the events A and B are such that
P(A) = 1/2 , P (A ∪ B) = 3/5 and
P(B) = p. Find p if they are
(i) mutually exclusive (ii) independent.

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{that}\mathrm{the}\mathrm{events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{such}\mathrm{that}\\ \mathrm{P}\left(\mathrm{A}\right)=\frac{1}{2},\mathrm{}\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\frac{3}{5} \mathrm{and}\mathrm{P}\left(\mathrm{B}\right)=\mathrm{p}\\ \left(\mathrm{i}\right)\mathrm{When}\mathrm{the}\mathrm{events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{mutually}\mathrm{exclusive}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=0\\ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{ }\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ \frac{3}{5}=\frac{1}{2}+\mathrm{p}-0\\ ⇒\mathrm{p}=\frac{3}{5}-\frac{1}{2}\\ =\frac{6-5}{10}\\ ⇒\mathrm{p}=\frac{1}{10}\\ \left(\mathrm{ii}\right)\mathrm{When}\mathrm{the}\mathrm{events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{independent}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right).\mathrm{ }\mathrm{P}\left(\mathrm{B}\right)\\ \therefore \mathrm{ }\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{ }\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ \mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{ }\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right).\mathrm{ }\mathrm{P}\left(\mathrm{B}\right)\\ ⇒\frac{3}{5}=\frac{1}{2}+\mathrm{p}-\frac{1}{2}×\mathrm{p}\\ ⇒ \frac{3}{5}-\frac{1}{2}=\mathrm{p}\left(1-\frac{1}{2}\right)\\ ⇒ \frac{6-5}{10}=\frac{1}{2}\mathrm{p}\\ \mathrm{p}=\frac{1}{10}×\frac{2}{1}=\frac{1}{5}\end{array}$

Q.8 Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find
(i) P(A ∩ B) (ii) P(A ∪ B)
(iii) P(A|B) (iv) P (B|A)

Ans

$\begin{array}{l}\mathrm{A}\mathrm{}\mathrm{and}\mathrm{}\mathrm{B}\mathrm{}\mathrm{be}\mathrm{}\mathrm{independent}\mathrm{}\mathrm{events}\mathrm{}\mathrm{with}\mathrm{P}\left(\mathrm{A}\right)=0.3\mathrm{}\mathrm{and}\mathrm{}\mathrm{P}\left(\mathrm{B}\right)=0.4\\ \left(\mathrm{i}\right) \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{ }=0.3×0.4\\ \mathrm{ }=0.12\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{ }=0.3+0.4-0.12\\ \mathrm{ }=0.7-0.12\\ \mathrm{ }=0.58\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{B}\right)}\\ \mathrm{ }=\frac{0.12}{0.4}\\ \mathrm{ }=0.3\\ \left(\mathrm{iv}\right)\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{A}\right)}\\ \mathrm{ }=\frac{0.12}{0.3}\\ \mathrm{ }=0.4\end{array}$

Q.9 If A and B are two events such that P(A) =1/4, P(B)= 1/2 and P(A∩B)= 1/8, find P(not A and not B).

Ans

$\begin{array}{l}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{two}\mathrm{events}\mathrm{such}\mathrm{that}\mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{1}}{4},\mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{1}}{2} \mathrm{and}\\ \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{\mathrm{1}}{8}\\ \mathrm{P}\left(\mathrm{not}\mathrm{ }\mathrm{A}\mathrm{and}\mathrm{not}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}‘\cap \mathrm{B}‘\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)‘\left[\begin{array}{l}\mathrm{By}\mathrm{De}\mathrm{Morgan}‘\mathrm{s}\mathrm{Law}\\ \left(\mathrm{A}‘\cap \mathrm{B}‘\right)=\left(\mathrm{A}\cup \mathrm{B}\right)‘\end{array}\right]\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)\\ \mathrm{ }=1-\left\{\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\right\}\\ \mathrm{ }=1-\left\{\frac{\mathrm{1}}{4}+\frac{\mathrm{1}}{2}-\frac{\mathrm{1}}{4}×\frac{\mathrm{1}}{2}\right\}\\ \mathrm{ }=1-\left(\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\right)\\ \mathrm{ }=1-\left(\frac{2+4-1}{8}\right)\\ \mathrm{ }=1-\frac{5}{8}\\ \mathrm{ }=\frac{8-5}{8}\mathrm{ }=\frac{3}{8}\\ \mathrm{Thus},\mathrm{ }\mathrm{P}\left(\mathrm{not}\mathrm{ }\mathrm{A}\mathrm{and}\mathrm{not}\mathrm{B}\right)\mathrm{ }\mathrm{is} \frac{3}{8}.\end{array}$

Q.10 Events A and B are such that P (A) =1/2, P(B)=7/12 and P(not A or not B) = 1/4. State whether A and B are independent?

Ans

$\begin{array}{l}\mathrm{Events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{such}\mathrm{that}\mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{1}}{2}, \mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{7}}{12} \mathrm{and}\\ \mathrm{P}\left(\mathrm{not}\mathrm{A}\mathrm{or}\mathrm{not}\mathrm{B}\right)\mathrm{}=\frac{\mathrm{1}}{4}\\ \mathrm{P}\left(\mathrm{not}\mathrm{ }\mathrm{A}\mathrm{or}\mathrm{not}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}‘\cup \mathrm{B}‘\right)\\ \mathrm{ }\frac{\mathrm{1}}{4}\mathrm{ }=\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)‘\left[\begin{array}{l}\mathrm{By}\mathrm{De}\mathrm{Morgan}‘\mathrm{s}\mathrm{Law}\\ \left(\mathrm{A}‘\cap \mathrm{B}‘\right)=\left(\mathrm{A}\cup \mathrm{B}\right)‘\end{array}\right]\\ \mathrm{ }\frac{\mathrm{1}}{4}=1-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=1-\frac{1}{4}\\ =\frac{3}{4}\\ \mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{1}}{2}×\frac{7}{12}\\ =\frac{7}{24}\\ \mathrm{So}, \mathrm{ }\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\ne \mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{Thus},\mathrm{A}\mathrm{}\mathrm{and}\mathrm{}\mathrm{B}\mathrm{}\mathrm{are}\mathrm{}\mathrm{independent}\mathrm{events}\mathrm{.}\end{array}$

Q.11 Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find
(i) P(A and B) (ii) P(A and not B)
(iii) P(A or B) (iv) P(neither A nor B)

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{two}\mathrm{independent}\mathrm{events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{such}\mathrm{that}\mathrm{P}\left(\mathrm{A}\right)=0.\mathrm{3},\mathrm{}\\ \mathrm{P}\left(\mathrm{B}\right)=0.\mathrm{6}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{A}\mathrm{and}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ =\mathrm{P}\left(\mathrm{A}\right).\mathrm{P}\left(\mathrm{B}\right)\\ =0.3×0.6\\ =0.18\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{A}\mathrm{and}\mathrm{not}\mathrm{B}\right)\\ =\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}‘\right)\\ =\mathrm{P}\left(\mathrm{A}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ =0.3-0.18\\ =0.12\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{A}\mathrm{or}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)\\ =\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\\ =0.3+0.6-0.18\\ =0.9-0.18\\ =0.72\\ \left(\mathrm{iv}\right)\mathrm{P}\left(\mathrm{neither}\mathrm{}\mathrm{A}\mathrm{}\mathrm{nor}\mathrm{}\mathrm{B}\right)\\ =\mathrm{P}\left(\mathrm{A}‘\cap \mathrm{B}‘\right)\\ =\left[\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)‘\right]\left[\begin{array}{l}\mathrm{By}\mathrm{De}\mathrm{Morgan}‘\mathrm{s}\mathrm{Law}\\ \mathrm{ }\mathrm{A}‘\cap \mathrm{B}‘=\left(\mathrm{A}\cup \mathrm{B}\right)‘\end{array}\right]\\ =1-\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)\\ =1-0.72\\ =0.28\end{array}$

Q.12 A die is tossed thrice. Find the probability of getting an odd number at least once.

Ans

$\begin{array}{l}\mathrm{ }\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{odd}\mathrm{number}\mathrm{in}\mathrm{a}\mathrm{throw}\mathrm{of}\mathrm{a}\mathrm{die}=\frac{3}{6}=\frac{1}{2}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{even}\mathrm{number}\mathrm{in}\mathrm{a}\mathrm{throw}\mathrm{of}\mathrm{a}\mathrm{die}=\frac{3}{6}=\frac{1}{2}\\ \mathrm{ }\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{even}\mathrm{number}\mathrm{three}\mathrm{times}=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\\ =\frac{1}{8}\\ \mathrm{Therefore},\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{odd}\mathrm{number}\mathrm{at}\mathrm{least}\mathrm{once}\\ =1-\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{odd}\mathrm{number}\mathrm{in}\mathrm{none}\mathrm{of}\mathrm{the}\mathrm{throws}\\ =\mathrm{}1-\mathrm{Probability}\mathrm{}\mathrm{of}\mathrm{}\mathrm{getting}\mathrm{}\mathrm{an}\mathrm{}\mathrm{even}\mathrm{}\mathrm{number}\mathrm{}\mathrm{thrice} \mathrm{ }=\mathrm{1}-\frac{1}{8}\\ \mathrm{ }=\frac{7}{8}\end{array}$

Q.13 Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.

Ans

$\begin{array}{l}\mathrm{Two}\mathrm{}\mathrm{balls}\mathrm{}\mathrm{are}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{at}\mathrm{}\mathrm{random}\mathrm{}\mathrm{with} \mathrm{replacement}\mathrm{}\mathrm{from}\mathrm{}\mathrm{a}\mathrm{}\mathrm{box}\mathrm{}\\ \mathrm{containing}\mathrm{}10\mathrm{}\mathrm{black} \mathrm{and}\mathrm{}8\mathrm{}\mathrm{red}\mathrm{}\mathrm{balls}.\\ \mathrm{ }\mathrm{Total}\mathrm{balls}=10+8=18\\ \mathrm{Black}\mathrm{balls}=10\\ \mathrm{ }\mathrm{Red}\mathrm{balls}=\mathrm{8}\\ \left(\mathrm{i}\right) \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{red}\mathrm{ball}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{draw}=\frac{8}{18}=\frac{4}{9}\\ \mathrm{ }\mathrm{The}\mathrm{ball}\mathrm{is}\mathrm{replaced}\mathrm{after}\mathrm{the}\mathrm{first}\mathrm{draw}.\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{second}\mathrm{red}\mathrm{ball}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{draw}=\frac{8}{18}=\frac{4}{9}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting} \mathrm{ }\mathrm{both}\mathrm{the}\mathrm{red}\mathrm{balls}=\frac{4}{9}×\frac{4}{9}\\ \mathrm{ }=\frac{16}{81}\\ \left(\mathrm{ii}\right) \mathrm{ }\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{black}\mathrm{ball}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{draw}=\frac{10}{18}\\ \mathrm{ }\mathrm{The}\mathrm{ball}\mathrm{is}\mathrm{replaced}\mathrm{after}\mathrm{the}\mathrm{first}\mathrm{draw}.\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{second}\mathrm{red}\mathrm{ball}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{draw}=\frac{8}{18}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{first}\mathrm{ball}\mathrm{as}\mathrm{black}\mathrm{and}\mathrm{second}\mathrm{ball}\mathrm{as}\mathrm{red}\\ =\frac{10}{18}×\frac{8}{18}\\ =\frac{20}{81}\\ \left(\mathrm{iii}\right) \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{first}\mathrm{ball}\mathrm{as}\mathrm{red}=\frac{8}{18}=\frac{4}{9}\\ \mathrm{The}\mathrm{ball}\mathrm{is}\mathrm{replaced}\mathrm{after}\mathrm{the}\mathrm{first}\mathrm{draw}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{second}\mathrm{ball}\mathrm{as}\mathrm{black}=\frac{10}{18}=\frac{5}{9}\\ \therefore \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{first}\mathrm{ball}\mathrm{as}\mathrm{red}\mathrm{and}\mathrm{ }\mathrm{second}\\ \mathrm{ball}\mathrm{as}\mathrm{black}=\frac{4}{9}×\frac{5}{9}\\ =\frac{20}{81}\\ \mathrm{Therefore},\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{first}\mathrm{ball}\mathrm{as}\mathrm{black}\mathrm{and}\mathrm{second}\mathrm{ball}\mathrm{as}\mathrm{red}=\frac{4}{9}×\frac{5}{9}\\ =\frac{20}{81}\\ \mathrm{Therefore},\mathrm{probability}\mathrm{that}\mathrm{one}\mathrm{of}\mathrm{them}\mathrm{is}\mathrm{black}\mathrm{and}\mathrm{other}\mathrm{is}\mathrm{red}\\ =\mathrm{}\frac{20}{81}+\frac{20}{81}\\ =\frac{40}{81}\end{array}$

Q.14 Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem.

Ans

$\begin{array}{l} \mathrm{Probability}\mathrm{of}\mathrm{solving}\mathrm{specific}\mathrm{problem}\mathrm{by}\mathrm{A},\mathrm{P}\left(\mathrm{A}\right)=\frac{1}{2}\\ \mathrm{ }\mathrm{Probability}\mathrm{of}\mathrm{solving}\mathrm{specific}\mathrm{problem}\mathrm{by}\mathrm{B},\mathrm{P}\left(\mathrm{B}\right)=\frac{1}{3}\\ \mathrm{Probability}\mathrm{of}\mathrm{not}\mathrm{solving}\mathrm{specific}\mathrm{problem}\mathrm{by}\mathrm{A},\mathrm{P}\left(\mathrm{A}‘\right)=1-\frac{1}{2}\\ =\frac{1}{2}\\ \mathrm{ }\mathrm{Probability}\mathrm{of}\mathrm{not}\mathrm{solving}\mathrm{specific}\mathrm{problem}\mathrm{by}\mathrm{B},\mathrm{P}\left(\mathrm{B}‘\right)=1-\frac{1}{3}\\ =\frac{2}{3}\\ \left(\mathrm{i}\right)\mathrm{Probability}\mathrm{of}\mathrm{Problem}\mathrm{solved}\mathrm{by}\mathrm{both}\mathrm{independetly},\\ \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\frac{1}{2}×\frac{1}{3}\\ =\frac{1}{6}\\ \mathrm{Probability}\mathrm{that}\mathrm{Problem}\mathrm{is}\mathrm{solved},\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)\\ =\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ =\frac{1}{2}+\frac{1}{3}-\frac{1}{2}×\frac{1}{3}\\ =\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\\ =\frac{3+2-1}{6}\\ =\frac{4}{6}=\frac{2}{3}\\ \left(\mathrm{ii}\right)\mathrm{Probability}\mathrm{that}\mathrm{exactly}\mathrm{one}\mathrm{of}\mathrm{them}\mathrm{solves}\mathrm{the}\mathrm{problem}\\ =\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}‘\right)+\mathrm{P}\left(\mathrm{B}\right)\mathrm{P}\left(\mathrm{A}‘\right)\\ =\frac{1}{2}×\frac{2}{3}+\frac{1}{3}×\frac{1}{2}\\ =\frac{1}{3}+\frac{1}{6}\\ =\frac{2+1}{6}\\ =\frac{3}{6}=\frac{1}{2}\end{array}$

Q.15 One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?
(i) E: ‘the card drawn is a spade’
F: ‘the card drawn is an ace’
(ii) E: ‘the card drawn is black’
F: ‘the card drawn is a king’
(iii) E: ‘the card drawn is a king or queen’
F: ‘the card drawn is a queen or jack’.

Ans

$\begin{array}{l}\mathrm{Total}\mathrm{cards}\mathrm{in}\mathrm{a}\mathrm{well}\mathrm{shuffled}\mathrm{deck}=52\\ \left(\mathrm{i}\right)\mathrm{}\mathrm{E}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{}\mathrm{a}\mathrm{}\mathrm{spade}’\\ \mathrm{P}\left(\mathrm{E}\right)=\frac{4}{52}\\ =\frac{1}{13}\\ \mathrm{}\mathrm{F}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{}\mathrm{an}\mathrm{}\mathrm{ace}’\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{13}{52}\mathrm{ }=\frac{1}{4}\\ \mathrm{ }\mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\frac{1}{52}\\ \mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)=\frac{1}{13}×\frac{1}{4}\\ \mathrm{ }=\frac{1}{52}\\ \therefore \mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)\\ \mathrm{Thus},\mathrm{ }\mathrm{the}\mathrm{events}\mathrm{E}\mathrm{and}\mathrm{F}\mathrm{are}\mathrm{independent}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{}\mathrm{E}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{}\mathrm{black}’\\ \mathrm{P}\left(\mathrm{E}\right)=\frac{13}{52}\\ =\frac{1}{4}\\ \mathrm{}\mathrm{F}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{}\mathrm{a}\mathrm{ }\mathrm{king}’\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{4}{52}\\ \mathrm{ }=\frac{1}{13}\\ \mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)=\frac{1}{4}×\frac{1}{13}\\ \mathrm{ }=\frac{1}{52}\\ \therefore \mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)=\mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)\\ \mathrm{Thus},\mathrm{ }\mathrm{the}\mathrm{events}\mathrm{E}\mathrm{and}\mathrm{F}\mathrm{are}\mathrm{independent}\mathrm{.}\\ \left(\mathrm{iii}\right)\mathrm{}\mathrm{E}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{a}\mathrm{king}\mathrm{}\mathrm{or}\mathrm{}\mathrm{queen}’\\ \mathrm{P}\left(\mathrm{E}\right)=\frac{8}{52}=\frac{2}{13}\\ \mathrm{}\mathrm{F}:‘\mathrm{the}\mathrm{}\mathrm{card}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{is}\mathrm{}\mathrm{a}\mathrm{ }\mathrm{queen}\mathrm{}\mathrm{or}\mathrm{}\mathrm{jack}’\\ \mathrm{P}\left(\mathrm{F}\right)=\frac{8}{52}\\ \mathrm{ }=\frac{2}{13}\\ \mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)=\frac{2}{13}×\frac{2}{13}\\ \mathrm{ }=\frac{4}{169}\\ \therefore \mathrm{P}\left(\mathrm{E}\cap \mathrm{F}\right)\ne \mathrm{P}\left(\mathrm{E}\right).\mathrm{ }\mathrm{P}\left(\mathrm{F}\right)\\ \mathrm{Thus},\mathrm{ }\mathrm{the}\mathrm{events}\mathrm{E}\mathrm{and}\mathrm{F}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{.}\end{array}$

Q.16 In a hostel, 60% of the students read Hindi news paper, 40% read English news paper and 20% read both Hindi and English news papers. A student is selected at random.
(a) Find the probability that she reads neither Hindi nor English news papers.
(b) If she reads Hindi news paper, find the probability that she reads English news paper.
(c) If she reads English news paper, find the probability that she reads Hindi news paper.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{H}=\mathrm{the}\mathrm{students}\mathrm{who}\mathrm{read}\mathrm{Hindi}\mathrm{newspaper}\mathrm{and}\\ \mathrm{E}=\mathrm{the}\mathrm{students}\mathrm{who}\mathrm{read}\mathrm{English}\mathrm{newspaper}.\\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that},\\ \mathrm{P}\left(\mathrm{H}\right)=60\mathrm{%}=\frac{60}{100}=0.6\\ \mathrm{P}\left(\mathrm{E}\right)=40\mathrm{%}=\frac{40}{100}=0.4\\ \mathrm{P}\left(\mathrm{H}\cap \mathrm{E}\right)=20\mathrm{%}=\frac{20}{100}=0.2\\ \left(\mathrm{i}\right)\mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{she}\mathrm{reads}\mathrm{neither}\mathrm{Hindi}\mathrm{nor}\mathrm{English}\mathrm{news}\mathrm{papers}=\mathrm{P}\left(\mathrm{H}\cup \mathrm{E}\right)‘\\ =1-\mathrm{P}\left(\mathrm{H}\cup \mathrm{E}\right)\\ =1-\left\{\mathrm{P}\left(\mathrm{H}\right)+\mathrm{P}\left(\mathrm{E}\right)-\mathrm{P}\left(\mathrm{H}\cap \mathrm{E}\right)\right\}\\ =1-\left(0.6+0.4-0.2\right)\\ =1-0.8\\ =0.2\\ \left(\mathrm{ii}\right)\mathrm{If}\mathrm{she}\mathrm{reads}\mathrm{Hindi}\mathrm{news}\mathrm{paper},\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{she}\mathrm{reads}\mathrm{English}\mathrm{news}\mathrm{paper}\\ =\mathrm{P}\left(\mathrm{E}|\mathrm{H}\right)\\ =\frac{\mathrm{P}\left(\mathrm{E}\cap \mathrm{H}\right)}{\mathrm{P}\left(\mathrm{H}\right)}\\ =\frac{0.2}{0.6}=\frac{1}{3}\\ \left(\mathrm{iii}\right)\mathrm{Probability}\mathrm{that}\mathrm{a}\mathrm{randomly}\mathrm{chosen}\mathrm{student}\mathrm{reads}\mathrm{Hindi}\mathrm{newspaper},\mathrm{if}\mathrm{she}\mathrm{reads}\mathrm{English}\mathrm{newspaper}\\ =\mathrm{P}\left(\mathrm{H}|\mathrm{E}\right)\\ =\frac{\mathrm{P}\left(\mathrm{H}\cap \mathrm{E}\right)}{\mathrm{P}\left(\mathrm{E}\right)}\\ =\frac{0.2}{0.4}=\frac{1}{2}\end{array}$

Q.17 The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
(A) 0 (B) 1/3 (C) 1/12 (D) 1/36

Ans

When two dice are rolled, the number of outcomes is 36.The only even prime number is 2. Let E be the event of getting an even prime number on each die.
So, E = {(2, 2)}
P (E) = (1/36)
Thus, the correct option is D.

Q.18 Two events A and B will be independent, if
(A) A and B are mutually exclusive
(B) P(A′B′) = [1 – P(A)][1 – P(B)]
(C) P(A) = P(B)
(D) P(A) + P(B) = 1

Ans

$\begin{array}{l}\mathrm{And},\mathrm{ }\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)=\mathrm{m}.\mathrm{n}\ne 0\\ \therefore \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\ne \mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{So},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{events}\mathrm{.}\\ \mathrm{Therefore},\mathrm{option}\mathrm{A}\mathrm{is}\mathrm{not}\mathrm{correct}\mathrm{.}\\ \left(\mathrm{B}\right)\mathrm{Two}\mathrm{events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{said}\mathrm{to}\mathrm{be}\mathrm{independent},\\ \mathrm{if}\mathrm{P}\left(\mathrm{AB}\right)=\mathrm{P}\left(\mathrm{A}\right)×\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{Consider}\mathrm{the}\mathrm{result}\mathrm{given}\mathrm{in}\mathrm{alternative}\mathrm{B}.\\ \mathrm{P}\left(\mathrm{A}‘\mathrm{B}‘\right)=\left[1-\mathrm{P}\left(\mathrm{A}\right)\right]\left[1-\mathrm{P}\left(\mathrm{B}\right)\right]\\ ⇒ \mathrm{ }\mathrm{P}\left(\mathrm{A}‘\cap \mathrm{B}‘\right)=1-\mathrm{P}\left(\mathrm{A}\right)-\mathrm{P}\left(\mathrm{B}\right)+\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ ⇒ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)‘=1-\mathrm{P}\left(\mathrm{A}\right)-\mathrm{P}\left(\mathrm{B}\right)+\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \left[\mathrm{By}\mathrm{De}\mathrm{â€‹}\mathrm{Morgan}‘\mathrm{s}\mathrm{Law}\right]\\ ⇒\mathrm{ }1-\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=1-\mathrm{P}\left(\mathrm{A}\right)-\mathrm{P}\left(\mathrm{B}\right)+\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ ⇒ \mathrm{ }\mathrm{P}\left(\mathrm{A}\cup \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ ⇒ \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{This}\mathrm{implies}\mathrm{that}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{independent},\mathrm{if}\\ \mathrm{P}\left(\mathrm{A}‘\mathrm{B}‘\right)=\left[1-\mathrm{P}\left(\mathrm{A}\right)\right]\left[1-\mathrm{P}\left(\mathrm{B}\right)\right]\\ \mathrm{Therefore},\mathrm{option}\mathrm{B}\mathrm{is}\mathrm{correct}\mathrm{.}\\ \left(\mathrm{C}\right)\mathrm{Let}\\ \mathrm{A}:\mathrm{Event}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{odd}\mathrm{number}\mathrm{on}\mathrm{throw}\mathrm{of}\mathrm{a}\mathrm{die}=\left\{1, 3, 5\right\}\\ \mathrm{B}:\mathrm{Event}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{even}\mathrm{number}\mathrm{on}\mathrm{throw}\mathrm{of}\mathrm{a}\mathrm{die}=\left\{2, 4, 6\right\}\\ \therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{3}{6}=\frac{1}{2}\mathrm{â€‹}\mathrm{and}\mathrm{P}\left(\mathrm{B}\right)=\frac{3}{6}=\frac{1}{2}\mathrm{â€‹}\\ \mathrm{So}, \mathrm{A}\cap \mathrm{B}=\mathrm{Ï•}⇒\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)=0\\ \mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}\ne 0\\ ⇒\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\ne \mathrm{ }\mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{So},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{events}\mathrm{.}\\ \mathrm{Therefore},\mathrm{option}\mathrm{C}\mathrm{is}\mathrm{not}\mathrm{correct}\mathrm{.}\\ \left(\mathrm{D}\right)\mathrm{If}\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)=1,\\ \mathrm{then},\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)\ne \mathrm{P}\left(\mathrm{A}\right)\mathrm{P}\left(\mathrm{B}\right)\\ \mathrm{So},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{not}\mathrm{independent}\mathrm{events}\mathrm{.}\\ \mathrm{Therefore},\mathrm{option}\mathrm{D}\mathrm{is}\mathrm{not}\mathrm{correct}\mathrm{.}\\ \mathrm{Thus},\mathrm{option}\mathbf{B}\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

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