Find the mean number of heads in three tosses of a fair coin.
Question: Find the mean number of heads in three tosses of a fair coin. Solution: Let X denote the success of getting heads. Therefore, the sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} It can be seen that X can take the value of 0, 1, 2, or 3. $\therefore \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{TTT})$ $=\mathrm{P}(\mathrm{T}) \cdot \mathrm{P}(\mathrm{T}) \cdot \mathrm{P}(\mathrm{T})$ $=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$ $=\frac{1}{8}$ $\therefore P(X=1)=P(H H ...
Read More →Solve the following equations:
Question: Solve the following equations: (i) $\tan x+\tan 2 x+\tan 3 x=0$ (ii) $\tan x+\tan 2 x=\tan 3 x$ (iii) $\tan 3 x+\tan x=2 \tan 2 x$ Solution: (i) We have: $\tan x+\tan 2 x+\tan 3 x=0$ Now, $\tan x+\tan 2 x+\tan (x+2 x)=0$ $\Rightarrow \tan x+\tan 2 x+\left(\frac{\tan x+\tan 2 x}{1-\tan x \tan 2 x}\right)=0$ $\Rightarrow(\tan x+\tan 2 x)(1-\tan x \tan 2 x)+\tan x+\tan 2 x=0$ $\Rightarrow(\tan x+\tan 2 x)(2-\tan x \tan 2 x)=0$ $\Rightarrow \tan x+\tan 2 x=0$ or $2-\tan x \tan 2 x=0$ Now, ...
Read More →The random variable X has probability distribution P(X) of the following form, where k is some number:
Question: The random variable X has probability distribution P(X) of the following form, wherekis some number: $\mathrm{P}(\mathrm{X})=\left\{\begin{array}{l}k, \text { if } x=0 \\ 2 k, \text { if } x=1 \\ 3 k, \text { if } x=2 \\ 0, \text { otherwise }\end{array}\right.$ (a) Determine the value of $k$. (b) Find $\mathrm{P}(\mathrm{X}2), \mathrm{P}(\mathrm{X} \geq 2), \mathrm{P}(\mathrm{X} \geq 2)$. Solution: (a) It is known that the sum of probabilities of a probability distribution of random v...
Read More →Solve the following equations:
Question: Solve the following equations: (i) $\tan x+\tan 2 x+\tan 3 x=0$ (ii) $\tan x+\tan 2 x=\tan 3 x$ (iii) $\tan 3 x+\tan x=2 \tan 2 x$ Solution: (i) We have: $\tan x+\tan 2 x+\tan 3 x=0$ Now, $\tan x+\tan 2 x+\tan (x+2 x)=0$ $\Rightarrow \tan x+\tan 2 x+\left(\frac{\tan x+\tan 2 x}{1-\tan x \tan 2 x}\right)=0$ $\Rightarrow(\tan x+\tan 2 x)(1-\tan x \tan 2 x)+\tan x+\tan 2 x=0$ $\Rightarrow(\tan x+\tan 2 x)(2-\tan x \tan 2 x)=0$ $\Rightarrow \tan x+\tan 2 x=0$ or $2-\tan x \tan 2 x=0$ Now, ...
Read More →find the value of
Question: If $a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, find the value of $a^{2}+b^{2}-5 a b$. Solution: According to question, $a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ $a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ $=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ $=\frac{(\sqrt{3})^{2}+(\sqrt{2})^{2}-2(\sqrt{3})(\sqrt{2})}{(\sqrt{...
Read More →A random variable X has the following probability distribution.
Question: A random variable X has the following probability distribution. Determine (i) $k$ (ii) $P(X3)$ (iii) $P(X6)$ (iv) $P(0X3)$ Solution: (i)It is known that the sum of probabilities of a probability distribution of random variables is one. $\therefore 0+k+2 k+2 k+3 k+k^{2}+2 k^{2}+\left(7 k^{2}+k\right)=1$ $\Rightarrow 10 k^{2}+9 k-1=0$ $\Rightarrow(10 k-1)(k+1)=0$ $\Rightarrow k=-1, \frac{1}{10}$ k= 1 is not possible as the probability of an event is never negative. $\therefore k=\frac{1}...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\frac{\cos \theta}{\operatorname{cosec} \theta+1}+\frac{\cos \theta}{\operatorname{cosec} \theta-1}=2 \tan \theta$ Solution: In the given question, we need to prove $\frac{\cos \theta}{\operatorname{cosec} \theta+1}+\frac{\cos \theta}{\operatorname{cosec} \theta-1}=2 \tan \theta$ Using the identity $a^{2}-b^{2}=(a+b)(a-b)$, we get $\frac{\cos \theta}{\operatorname{cosec} \theta+1}+\frac{\cos \theta}{\operatorname{cosec} \theta-1}=\frac{\co...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $1+\frac{\cot ^{2} \theta}{1+\operatorname{cosec} \theta}=\operatorname{cosec} \theta$ Solution: In the given question, we need to prove $1+\frac{\cot ^{2} \theta}{1+\operatorname{cosec} \theta}=\operatorname{cosec} \theta$ Using $\cot \theta=\frac{\cos \theta}{\sin \theta}$ and $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$, we get $1+\frac{\cot ^{2} \theta}{1+\operatorname{cosec} \theta}=\frac{1+\operatorname{cosec} \theta+\cot ^{2} ...
Read More →Solve the following equations:
Question: Solve the following equations: (i) $\cos x+\cos 2 x+\cos 3 x=0$ (ii) $\cos x+\cos 3 x-\cos 2 x=0$ (iii) $\sin x+\sin 5 x=\sin 3 x$ (iv) $\cos x \cos 2 x \cos 3 x=\frac{1}{4}$ (v) $\cos x+\sin x=\cos 2 x+\sin 2 x$ (vi) $\sin x+\sin 2 x+\sin 3=0$ (vii) $\sin x+\sin 2 x+\sin 3 x+\sin 4 x=0$ (viii) $\sin 3 x-\sin x=4 \cos ^{2} x-2$ (ix) $\sin 2 x-\sin 4 x+\sin 6 x=0$ Solution: (i) $\cos x+\cos 2 x+\cos 3 x=0$ Now, $(\cos x+\cos 3 x)+\cos 2 x=0$ $\Rightarrow 2 \cos \left(\frac{4 x}{2}\right...
Read More →Solve the following equations:
Question: Solve the following equations: (i) $\cos x+\cos 2 x+\cos 3 x=0$ (ii) $\cos x+\cos 3 x-\cos 2 x=0$ (iii) $\sin x+\sin 5 x=\sin 3 x$ (iv) $\cos x \cos 2 x \cos 3 x=\frac{1}{4}$ (v) $\cos x+\sin x=\cos 2 x+\sin 2 x$ (vi) $\sin x+\sin 2 x+\sin 3=0$ (vii) $\sin x+\sin 2 x+\sin 3 x+\sin 4 x=0$ (viii) $\sin 3 x-\sin x=4 \cos ^{2} x-2$ (ix) $\sin 2 x-\sin 4 x+\sin 6 x=0$ Solution: (i) $\cos x+\cos 2 x+\cos 3 x=0$ Now, $(\cos x+\cos 3 x)+\cos 2 x=0$ $\Rightarrow 2 \cos \left(\frac{4 x}{2}\right...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\frac{1-\tan ^{2} A}{\cot ^{2} A-1}=\tan ^{2} A$ Solution: In the given question, we need to prove $\frac{1-\tan ^{2} A}{\cot ^{2} A-1}=\tan ^{2} A$ Now, using $\tan \theta=\frac{\sin \theta}{\cos \theta}$ and $\cot \theta=\frac{\cos \theta}{\sin \theta}$ in the L.H.S, we get $\frac{1-\tan ^{2} A}{\cot ^{2} A-1}=\frac{1-\frac{\sin ^{2} A}{\cos ^{2} A}}{\frac{\cos ^{2} A}{\sin ^{2} A}-1}$ $=\frac{\frac{\cos ^{2} A-\sin ^{2} A}{\cos ^{2} A}}...
Read More →A coin is biased so that the head is 3 times as likely to occur as tail.
Question: A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails. Solution: Let the probability of getting a tail in the biased coin bex. $\therefore P(T)=x$ $\Rightarrow P(H)=3 x$ For a biased coin, P (T) + P (H) = 1 $\Rightarrow x+3 x=1$ $\Rightarrow 4 x=1$ $\Rightarrow x=\frac{1}{4}$ $\therefore \mathrm{P}(\mathrm{T})=\frac{1}{4}$ and $\mathrm{P}(\mathrm{H})=\frac{3}{4}$ When the coin is tossed...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\tan ^{2} A+\cot ^{2} A=\sec ^{2} A \operatorname{cosec}^{2} A-2$ Solution: In the given question, we need to prove $\tan ^{2} A+\cot ^{2} A=\sec ^{2} A \operatorname{cosec}^{2} A-2$ Now, using $\tan \theta=\frac{\sin \theta}{\cos \theta}$ and $\cot \theta=\frac{\cos \theta}{\sin \theta}$ in L.H.S, we get $\tan ^{2} A+\cot ^{2} A=\frac{\sin ^{2} A}{\cos ^{2} A}+\frac{\cos ^{2} A}{\sin ^{2} A}$ $=\frac{\sin ^{4} A+\cos ^{4} A}{\cos ^{2} A \...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}$ Solution: In the given question, we need to prove $\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}$. Here, we will first solve the L.H.S. Now, using $\sec \theta=\frac{1}{\cos \theta}$ and $\tan \theta=\frac{\sin \theta}{\cos \theta}$, we get $\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\left(\frac{1}{\cos A}+\frac{\si...
Read More →From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement..
Question: From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs. Solution: It is given that out of 30 bulbs, 6 are defective. ⇒ Number of non-defective bulbs = 30 6 = 24 4 bulbs are drawn from the lot with replacement. Let X be the random variable that denotes the number of defective bulbs in the selected bulbs. $\therefore P(X=0)=P(4$ non-defective and 0 defective $)={ }^{4} ...
Read More →show that
Question: If $a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ and $b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$, show that $3 a^{2}+4 a b-3 b^{2}=4+\frac{56}{3} \sqrt{10}$. Solution: According to question, $a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ and $b=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ $a=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ $=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}} \times \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ $=\frac{(\sqrt{5}+\sqrt{2})^{2}}{(\sqrt{5})^{2}...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. (i) $\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}=\frac{1+\sin \theta}{\cos \theta}$ (ii) $\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$ (iii) $\frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1}=\operatorname{cosec} \theta+\cot \theta$ (iv) $(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)=\sec \theta+\operatorname{cosec} \theta$ Solution: (i) We have to prove the fo...
Read More →Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
Question: Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as (i)number greater than 4 (ii)six appears on at least one die Solution: When a die is tossed two times, we obtain (6 6) = 36 number of observations. Let X be the random variable, which represents the number of successes. i. Here, success refers to the number greater than 4 . $P(X=0)=P$ (number less than or equal to 4 on both the tosses) $=\frac{4}{6} \times \frac{4}{6}=\fra...
Read More →Solve the following equations:
Question: Solve the following equations: (i) $\sin ^{2} x-\cos x=\frac{1}{4}$ (ii) $2 \cos ^{2} x-5 \cos x+2=0$ (iii) $2 \sin ^{2} x+\sqrt{3} \cos x+1=0$ (iv) $4 \sin ^{2} x-8 \cos x+1=0$ (v) $\tan ^{2} x+(1-\sqrt{3}) \tan x-\sqrt{3}=0$ (vi) $3 \cos ^{2} x-2 \sqrt{3} \sin x \cos x-3 \sin ^{2} x=0$ (vii) $\cos 4 x=\cos 2 x$ Solution: (i) $\sin ^{2} x-\cos x=\frac{1}{4}$ $\Rightarrow 1-\cos ^{2} x-\cos x=\frac{1}{4}$ $\Rightarrow 4-4 \cos ^{2} x-4 \cos x=1$ $\Rightarrow 4 \cos ^{2} x+6 \cos x-2 \c...
Read More →Find the probability distribution of
Question: Find the probability distribution of (i)number of heads in two tosses of a coin (ii)number of tails in the simultaneous tosses of three coins (iii)number of heads in four tosses of a coin Solution: (i)When one coin is tossed twice, the sample space is {HH, HT, TH, TT} Let X represent the number of heads. X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0 Therefore, X can take the value of 0, 1, or 2. It is known that, $\mathrm{P}(\mathrm{HH})=\mathrm{P}(\mathrm{HT})=\mathrm{P}(\mathrm{TH})...
Read More →show that
Question: If $x=\frac{5-\sqrt{3}}{5+\sqrt{3}}$ and $y=\frac{5+\sqrt{3}}{5-\sqrt{3}}$, show that $x^{2}-y^{2}=-\frac{10 \sqrt{3}}{11}$ Solution: Disclaimer: The question is incorrect. $x=\frac{5-\sqrt{3}}{5+\sqrt{3}}$ $\Rightarrow x=\frac{5-\sqrt{3}}{5+\sqrt{3}} \times \frac{5-\sqrt{3}}{5-\sqrt{3}}$ $\Rightarrow x=\frac{(5-\sqrt{3})^{2}}{5^{2}-(\sqrt{3})^{2}}$ $\Rightarrow x=\frac{25+3-10 \sqrt{3}}{25-3}$ $\Rightarrow x=\frac{28-10 \sqrt{3}}{22}$ $\Rightarrow x=\frac{14-5 \sqrt{3}}{11}$ $y=\frac{...
Read More →Let X represents the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times.
Question: Let X represents the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X? Solution: A coin is tossed six times and X represents the difference between the number of heads and the number of tails. $\therefore \mathrm{X}(6 \mathrm{H}, 0 \mathrm{~T})=|6-0|=6$ $X(5 H, 1 T)=|5-1|=4$ $X(4 \mathrm{H}, 2 T)=|4-2|=2$ $X(3 H, 3 T)=|3-3|=0$ $X(2 H, 4 T)=|2-4|=2$ $X(1 \mathrm{H}, 5 \mathrm{~T})=|1-5|=4$ $\mathrm{X}(\m...
Read More →find the value of
Question: If $x=2+\sqrt{3}$, find the value of $x^{3}+\frac{1}{x^{3}}$. Solution: $x=2+\sqrt{3}$ ..........(1) $\Rightarrow \frac{1}{x}=\frac{1}{2+\sqrt{3}}$ $\Rightarrow \frac{1}{x}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$ $\Rightarrow \frac{1}{x}=\frac{2-\sqrt{3}}{2^{2}-(\sqrt{3})^{2}}$ $\Rightarrow \frac{1}{x}=\frac{2-\sqrt{3}}{4-3}$ $\Rightarrow \frac{1}{x}=2-\sqrt{3}$ ....(2) Adding (1) and (2), we get $x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4 \quad \ldots(3)$ Cubing both sid...
Read More →Find the general solutions of the following equations:
Question: Find the general solutions of the following equations: (i) $\sin 2 x=\frac{\sqrt{3}}{2}$ (ii) $\cos 3 x=\frac{1}{2}$ (iii) $\sin 9 x=\sin \mathrm{x}$ (iv) $\sin 2 x=\cos 3 x$ (v) $\tan x+\cot 2 x=0$ (vi) $\tan 3 x=\cot x$ (vii) $\tan 2 x \tan x=1$ (viii) $\tan m x+\cot n x=0$ (ix) $\tan p x=\cot q x$ (x) $\sin 2 x+\cos x=0$ (xi) $\sin x=\tan x$ (xii) $\sin 3 x+\cos 2 x=0$ Solution: We have: (i) $\sin 2 x=\frac{\sqrt{3}}{2}$ $\Rightarrow \sin 2 x=\sin \frac{\pi}{3}$ $\Rightarrow 2 x=n \...
Read More →An urn contains 5 red and 2 black balls.
Question: An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represents the number of black balls. What are the possible values of X? Is X a random variable? Solution: The two balls selected can be represented as BB, BR, RB, RR, where B represents a black ball and R represents a red ball. X represents the number of black balls. X (BB) = 2 X (BR) = 1 X (RB) = 1 X (RR) = 0 Therefore, the possible values of X are 0, 1, and 2. Yes, X is a random variable....
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