Probability

$A^2=\left(\begin{array}{ll}\mathrm{c}\mathrm{o}\mathrm{s}?2\theta & i\mathrm{s}\mathrm{i}\mathrm{n}?2\theta \\ i\mathrm{s}\mathrm{i}\mathrm{n}?2\theta & \mathrm{c}\mathrm{o}\mathrm{s}?2\theta \end{array}\right)$

Similarly, $A^5=\left(\begin{array}{cc}\mathrm{c}\mathrm{o}\mathrm{s}?5\theta & i\mathrm{s}\mathrm{i}\mathrm{n}?5\theta \\ i\mathrm{s}\mathrm{i}\mathrm{n}?5\theta & \mathrm{c}\mathrm{o}\mathrm{s}?5\theta \end{array}\right)=\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)$

(1) $a^2+b^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta -{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =\mathrm{c}\mathrm{o}\mathrm{s}?10\theta =\mathrm{c}\mathrm{o}\mathrm{s}?{75}^{?}$

(2) $a^2-d^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta -{\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta =0$

(3) $a^2-b^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta +{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =1$

(4) $a^2-c^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta +{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =1$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}WeknowthatE\left(X\right)={\displaystyle \sum _{i=1}^n{X}_i{P}_i}\\ =\left(-4\right)\left(0.1\right)+\left(-3\right)\left(0.2\right)+\left(-2\right)\left(0.3\right)+\left(-1\right)\left(0.2\right)+0\left(0.2\right)\\ =-0.4-0.6-0.6-0.2=-1.8\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat:Bagcontains5redand3blueballs.\\ ofgettingexactlyoneredballif3ballsarerandomlydrawnwithoutreplacement.\\ P\left(R\right).P\left(B\right).P\left(B\right)+P\left(B\right).P\left(R\right).P\left(B\right)+P\left(B\right).P\left(B\right).P\left(R\right)\\ =\frac{5}{8}.\frac{3}{7}.\frac{2}{6}+\frac{3}{8}.\frac{5}{7}.\frac{2}{6}+\frac{3}{8}.\frac{2}{7}.\frac{5}{6}=\frac{30}{336}+\frac{30}{336}+\frac{30}{336}=\frac{90}{336}=\frac{15}{56}\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Sol:

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Foraprobabilitydistribution,weknowthatif{P}_i\ge 0\\ \left(i\right){\displaystyle \sum _{i=1}^n{P}_i=1\Rightarrow k+k^2+}2k^2+k=1\\ \Rightarrow 3k^2+2k-1=0\Rightarrow 3k^2+3k-k-1=0\\ \Rightarrow 3k\left(k+1\right)-1\left(k+1\right)=0\Rightarrow \left(3k-1\right)\left(k+1\right)=0\\ \therefore k=\frac{1}{3}andk=-1\\ Butk\ge 0\therefore k=\frac{1}{3}\\ \left(ii\right)Meanofthedistribution\\ E\left(X\right)={\displaystyle \sum _{i=1}^n{X}_i{P}_i}=0.5k+1.5k^2+1.5\left(2k^2\right)+2k\\ =\frac{k}{2}+k^2+3k^2+2k=4k^2+\frac{5}{2}k\\ =4{\left(\frac{1}{3}\right)}^2+\frac{5}{2}\left(\frac{1}{3}\right)=\frac{4}{9}+\frac{5}{6}=\frac{23}{18}\end{array}$

95.

$\begin{array}{l}P\left(A\right)+P\left(B\right)-P\left(AandB\right)=P\left(A\right)\\ \Rightarrow P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=P\left(A\right)\\ \Rightarrow P\left(B\right)-P\left(A\cap B\right)=0\\ \Rightarrow P\left(A\cap B\right)=P\left(B\right)\\ \therefore P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{P\left(B\right)}{P\left(B\right)}=1\end{array}$

Therefore, option (B) is correct.

94.

$\begin{array}{l}P\left(A|B\right)>P\left(A\right)\\ \Rightarrow \frac{P\left(A\cap B\right)}{P\left(B\right)}>P\left(A\right)\\ \Rightarrow P\left(A\cap B\right)>P\left(A\right).P\left(B\right)\\ \Rightarrow \frac{P\left(A\cap B\right)}{P\left(A\right)}>P\left(B\right)\\ \Rightarrow P\left(B|A\right)>P\left(B\right)\end{array}$

Therefore, option (C) is correct.

93. $P\left(A\right)\ne 0$ and $P\left(B|A\right)=1$

$\begin{array}{l}P\left(B\cap A\right)=\frac{P\left(B\cap A\right)}{P\left(A\right)}\\ 1=\frac{P\left(B\cap A\right)}{P\left(A\right)}\\ P\left(A\right)=P\left(B\cap A\right)\\ \Rightarrow A\subset B\end{array}$

Therefore, option (A) is correct.

92. Let E₁ = Ball transferred from Bag I to Bag II is red

E₂ = Ball transferred from Bag I to Bag Ii is black

A = Ball drawn from Bag II is red in colour

P (E₁) =3/7 and P (E₂) = 4/7

Let A be the event that the ball drawn is red.

When a red ball is transferred from bag I to II,

P (A | E₁) = 5/10 = 1/2

When a black ball is transferred from bag I to II,

P (A | E₂) = 4/10 = 2/5

$\begin{array}{l}\therefore P\left(E_2|A\right)=\frac{P\left(E_2\right)P\left(A|E_2\right)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)}\\ =\frac{\frac{4}{7}*\frac{2}{5}}{\frac{3}{7}*\frac{1}{2}+\frac{4}{7}*\frac{2}{5}}\\ =\frac{16}{31}\end{array}$

91. Let the event in which A fails and B fails be denoted by E_A and E_B.

P (EA) = 0.2

P (E_A ∩ E_B) = 0.15

P (B fails alone) = P (E_B) − P (E_A ∩ E_B)

⇒ 0.15 = P (E_B) − 0.15

⇒ P (E_B) = 0.3

$P\left(E_A|E_B\right)=\frac{P\left(E_A\cap E_B\right)}{P\left(E_B\right)}=\frac{0.15}{0.3}=0.5$

(ii) P (A fails alone) = P (E_A) − P (E_A ∩ E_B)

= 0.2 − 0.15

= 0.05