The probability of throwing a number greater than 2 with a fair dice is

Question: The probability of throwing a number greater than 2 with a fair dice is (a) $\frac{3}{5}$ (b) $\frac{2}{5}$ (c) $\frac{2}{3}$ (d) $\frac{1}{3}$ Solution: GIVEN: A dice is thrown once TO FIND: Probability of getting a number greater than 2. Total number on a dice is 6. Number greater than 2 is 3, 4, 5 and 6 Total number greater than 2 is 4 We know that PROBABILITY = Hence probability of getting a number greater than 2 is equal to Hence the correct option is...

Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from

Question: Show that the ratio of the sum of first $n$ terms of a G.P. to the sum of terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term is $\frac{1}{r^{n}}$. Solution: Letabe the first term andrbe the common ratio of the G.P. Sum of the first $n$ terms of the series $=a_{1}+a_{2}+a_{3}+\ldots+a_{n}$ Similarly, sum of the terms from $(n+1)^{\text {th }}$ to $2 n^{\text {th }}$ term $=a_{n+1}+a_{n+2}+\ldots+a_{2 n}$ $\therefore$ Required ratio $=\frac{a_{1}+a_{2}+a_{3}+\ldots+a_{n}}{a_{...

If a number x is chosen from the numbers 1, 2, 3,

Question: If a numberxis chosen from the numbers 1, 2, 3, and a numberyis selected from the numbers 1, 4, 9. Then,P(xy 9) (a) $\frac{7}{9}$ (b) $\frac{5}{9}$ (c) $\frac{2}{3}$ (d) $\frac{1}{9}$ Solution: GIVEN:xis chosen from the numbers 1, 2, 3 and y is chosen from the numbers 1, 4, 9 TO FIND: Probability of getting We will make multiplication table forxandysuch that $x y=\mid \times 1=1$ $=1 \times 4=4$ $=1 \times 9=9$ $=2 \times 1=2$ $=2 \times 4=8$ $=2 \times 9=18$ $=3 \times 1=3$ $=3 \times...

Solve the following

Question: If $S_{1}, S_{2}, S_{3}$ be respectively the sums of $n, 2 n, 3 n$ terms of a G.P., then prove that $S_{1}^{2}+S_{2}^{2}=S_{1}\left(S_{2}+S_{3}\right)$. Solution: Letabe the first term andrbe the common ratio of the given G.P. Sum of $n$ terms, $S_{1}=a\left(\frac{r^{n}-1}{r-1}\right)$ ...(1) Sum of $2 n$ terms, $S_{2}=a\left(\frac{r^{2 n}-1}{r-1}\right)$ $\Rightarrow S_{2}=a\left[\frac{\left(r^{n}\right)^{2}-1^{2}}{r-1}\right]$ $\Rightarrow S_{2}=a\left[\frac{\left(r^{n}-1\right)\left...

A number is selected at random from the numbers 3, 5, 5, 7, 7, 7, 9, 9, 9, 9

Question: A number is selected at random from the numbers 3, 5, 5, 7, 7, 7, 9, 9, 9, 9 The probability that the selected number is their average is (a) $\frac{1}{10}$ (b) $\frac{3}{10}$ (c) $\frac{7}{10}$ (d) $\frac{9}{10}$ Solution: GIVEN: A number is selected from the numbers 3,5,5,7,7,7,9,9,9,9 TO FIND: Probability that the selected number is the average of the numbers Total numbers are 10 Average of numbers $=\frac{3+5+5+7+7+7+9+9+9+9}{10}$ $=\frac{70}{10}$ $=7$ Total numbers of numbers whic...

A bag contains three green marbles,

Question: A bag contains three green marbles, four blue marbles, and two orange marbles, If a marble is picked at random, then the probability that it is not an orange marble is (a) $\frac{1}{4}$ (b) $\frac{1}{3}$ (C) $\frac{4}{9}$ (d) $\frac{7}{9}$ Solution: GIVEN: A bag contains 3green, 4blue and 2orange marbles TO FIND: Probability of not getting an orange marble Total number of balls 3+4+2=9 Total number of non orange marbles that is green and blue balls is We know that PROBABILITY = Hence p...

The probability of guessing the correct answer to a certain test questions is

Question: The probability of guessing the correct answer to a certain test questions is $\frac{x}{12}$. If the probability of not guessing the correct answer to this question is $\frac{2}{3}$, then $x=$ (a) 2(b) 3(c) 4(d) 6 Solution: GIVEN: Probability of guessing a correct answer to a certain question is Probability of not guessing a correct answer to a same question TO FIND: The value ofx CALCULATION:We know that sum of probability of occurrence of an event and probability of non occurrence of...

The fifth term of a G.P. is 81 whereas its second term is 24.

Question: The fifth term of a G.P. is 81 whereas its second term is 24. Find the series and sum of its first eight terms. Solution: Letabe the first term andrbe the common ratio of the G.P. $a_{2}=24$ $\Rightarrow a r^{2-1}=24$ $\Rightarrow a r=24$ ....(i) Similarly, $a_{5}=81$ $\Rightarrow a r^{5-1}=24$ $\Rightarrow a r^{4}=81$ $\Rightarrow \frac{24 \times r^{4}}{r}=81$ [From (i)] $\Rightarrow r^{3}=\frac{81}{24}$ $\therefore r^{3}=\frac{27}{8}$ $\Rightarrow r=\frac{3}{2}$ Putting $r=\frac{3}{2...

A number x is chosen at random from the numbers −3, −2, −1, 0, 1, 2, 3 the probability that | x | < 2 is

Question: A number x is chosen at random from the numbers 3, 2, 1, 0, 1, 2, 3 the probability that | x | 2 is (a) $\frac{5}{7}$ (b) $\frac{2}{7}$ (c) $\frac{3}{7}$ (d) $\frac{1}{7}$ Solution: GIVEN: A numberxis chosen from the numbers 3, 2, 1, 0, 1, 2 and 3 TO FIND: Probability of getting Total numbers are 7 Number $x$ such that $|x|2$ are $-1,0,1$ Total numbers $x$ such $|x|2$ are 3 We know that PROBABILITY $=\frac{\text { Number of favourable event }}{\text { Total number of event }}$ Hence th...

Question: If $[1-1 x]\left[\begin{array}{rrr}0 1 -1 \\ 2 1 3 \\ 1 1 1\end{array}\right]\left[\begin{array}{l}0 \\ 1 \\ 1\end{array}\right]=0$, find $x$ Solution: Given : $\left[\begin{array}{lll}1 -1 x\end{array}\right]\left[\begin{array}{ccc}0 1 -1 \\ 2 1 3 \\ 1 1 1\end{array}\right]\left[\begin{array}{l}0 \\ 1 \\ 1\end{array}\right]=0$ $\Rightarrow\left[\begin{array}{lll}0-2+x 1-1+x -1-3+x\end{array}\right]\left[\begin{array}{l}0 \\ 1 \\ 1\end{array}\right]=0$ $\Rightarrow\left[\begin{array}{l...

In a single throw of a die, the probability of getting a multiple of 3 is

Question: In a single throw of a die, the probability of getting a multiple of 3 is (a) $\frac{1}{2}$ (b) $\frac{1}{3}$ (c) $\frac{1}{6}$ (d) $\frac{2}{3}$ Solution: GIVEN: A dice is thrown once TO FIND: Probability of getting a multiple of 3. Total number on a dice is 6. Numbers which are on multiple of 3 are 3 and 6 Total number of numbers which are multiple of 3 is 2 We know that PROBABILITY = Hence probability of getting a number which are multiple of 3 is equal to Hence the correct option i...

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Question: If $[x41]\left[\begin{array}{rrr}2 1 2 \\ 1 0 2 \\ 0 2 -4\end{array}\right]\left[\begin{array}{r}x \\ 4 \\ -1\end{array}\right]=0$, find $x$ Solution: Given : $\left[\begin{array}{lll}x 4 1\end{array}\right]\left[\begin{array}{ccc}2 1 2 \\ 1 0 2 \\ 0 2 -4\end{array}\right]\left[\begin{array}{c}x \\ 4 \\ -1\end{array}\right]=0$ $\Rightarrow\left[\begin{array}{lll}2 x+4+0 x+0+2 2 x+8-4\end{array}\right]\left[\begin{array}{c}x \\ 4 \\ -1\end{array}\right]=0$ $\Rightarrow\left[\begin{array...

If three coins are tossed simultaneously, then the probability of getting at least two heads, is

Question: If three coins are tossed simultaneously, then the probability of getting at least two heads, is (a) $\frac{1}{4}$ (b) $\frac{3}{8}$ (C) $\frac{1}{2}$ (d) $\frac{1}{4}$ Solution: GIVEN: Three coins are tossed simultaneously. TO FIND: Probability of getting at least two head. When three coins are tossed then the outcome will be TTT, THT, TTH, THH. HTT, HHT, HTH, HHH Hence total number of outcome is 8. At least two heads means that, THH, HHT, HTH and HHH are favorable events Hence total ...

if

Question: (i) If $\left[\begin{array}{lll}1 1 x]\end{array}\left[\begin{array}{lll}1 0 2 \\ 0 2 1 \\ 2 1 0\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=0\right.$, find $x$. (ii) If $\left[\begin{array}{ll}2 3 \\ 5 7\end{array}\right]\left[\begin{array}{cc}1 -3 \\ -2 4\end{array}\right]=\left[\begin{array}{cc}-4 6 \\ -9 x\end{array}\right]$, find $x$. Solution: (i) Given : $\left[\begin{array}{lll}1 1 x\end{array}\right]\left[\begin{array}{lll}1 0 2 \\ 0 2 1 \\ 2 1 0\end{a...

In Q. No. 1, the probability that the digit is a multiple of 3 is

Question: In Q. No. 1, the probability that the digit is a multiple of 3 is (a) $\frac{1}{3}$ (b) $\frac{2}{3}$ (c) $\frac{1}{9}$ (d) $\frac{2}{9}$ Solution: GIVEN: digits are chosen from 1, 2, 3, 4, 5, 6, 7, 8, 9 are placed in a box and mixed thoroughly. One digit is picked at random. TO FIND: Probability of getting a multiple of 3 Total number of digits is 9 Digits that are multiple of 3 are 3, 6 and 9 Total digits that are multiple of 3 are 3 We know that PROBABILITY = Hence probability of ge...

Find the sum :

Question: Find the sum : $\sum_{n=1}^{10}\left[\left(\frac{1}{2}\right)^{n-1}+\left(\frac{1}{5}\right)^{n+1}\right]$. Solution: $\sum_{n=1}^{10}\left[\left(\frac{1}{2}\right)^{n-1}+\left(\frac{1}{5}\right)^{n+1}\right]$ $=\sum_{n=1}^{10}\left(\frac{1}{2}\right)^{n-1}+\sum_{n=1}^{10}\left(\frac{1}{5}\right)^{n+1}$ $=\left\{1+\frac{1}{2}+\frac{1}{4}+\ldots+\left(\frac{1}{2}\right)^{9}\right\}+\left\{\frac{1}{5^{2}}+\frac{1}{5^{3}}+\frac{1}{5^{4}}+\ldots+\frac{1}{5^{11}}\right\}$ $=1\left(\frac{1-\...

In Q. No. 1, The probability that the digit is even is

Question: In Q. No. 1, The probability that the digit is even is (a) $\frac{4}{9}$ (b) $\frac{5}{9}$ (c) $\frac{1}{9}$ (d) $\frac{2}{3}$ Solution: GIVEN: digits are chosen from 1, 2, 3, 4, 5, 6, 7, 8, 9 are placed in a box and mixed thoroughly. One digit is picked at random. TO FIND: Probability of getting an even digit Total number of digits is 9 Digits that are even number are 2, 4, 6, and 8 Total number of even digits is 4. We know that PROBABILITY = Hence probability of getting an even digit...

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Question: If $A=\left[\begin{array}{rrr}4 -1 -4 \\ 3 0 -4 \\ 3 -1 -3\end{array}\right]$, show that $A^{2}=/ 3$ Solution: Here, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{ccc}4 -1 -4 \\ 3 0 -4 \\ 3 -1 -3\end{array}\right]\left[\begin{array}{ccc}4 -1 -4 \\ 3 0 -4 \\ 3 -1 -3\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ccc}16-3-12 -4+0+4 -16+4+12 \\ 12+0-12 -3+0+4 -12+0+12 \\ 12-3-9 -3+0+3 -12+4+9\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{lll}1 0 0 \\ 0 1 0 \\ 0...

Mark the correct alternative in each of the following:

Question: Mark the correct alternative in each of the following:If a digit is chosen at random from the digit 1, 2, 3, 4, 5, 6, 7, 8, 9, then the probability that it is odd, is (a) $\frac{4}{9}$ (b) $\frac{5}{9}$ (c) $\frac{1}{9}$ (d) $\frac{2}{3}$ Solution: GIVEN: digits are chosen from 1, 2, 3 4, 5, 6, 7, 8, 9 are placed in a box and mixed thoroughly. One digit is picked at random. TO FIND: Probability of getting an odd digit Total number of digits is 9 Digit that are odd number are 1,3,5,7,9 ...

The 4th and 7th terms of a G.P.

Question: The 4 th and 7 th terms of a G.P. are $\frac{1}{27}$ and $\frac{1}{729}$ respectively. Find the sum of $n$ terms of the G.P. Solution: Letabe the first term andrbe the common ratio of the G.P. $\therefore a_{4}=\frac{1}{27}$ $\Rightarrow a r^{4-1}=\frac{1}{27}$ $\Rightarrow a r^{3}=\frac{1}{27}$ $\Rightarrow\left(a r^{3}\right)^{2}=\frac{1}{27^{2}}$ $\Rightarrow a^{2} r^{6}=\frac{1}{729}$ $\Rightarrow a r^{6}=\frac{1}{729 a}$ ...(i) Similarly, $\mathrm{a}_{7}=\frac{1}{729}$ $\Rightarro...

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Question: If $A=\left[\begin{array}{rrr}2 -3 -5 \\ -1 4 5 \\ 1 -3 -4\end{array}\right]$, show that $A^{2}=A$ Solution: Here, $A^{2}=A A$ $\Rightarrow A^{2}=\left[\begin{array}{ccc}2 -3 -5 \\ -1 4 5 \\ 1 -3 -4\end{array}\right]\left[\begin{array}{ccc}2 -3 -5 \\ -1 4 5 \\ 1 -3 -4\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ccc}4+3-5 -6-12+15 -10-15+20 \\ -2-4+5 3+16-15 5+20-20 \\ 2+3-4 -3-12+12 -5-15+16\end{array}\right]$ $\Rightarrow A^{2}=\left[\begin{array}{ccc}2 -3 -5 \\ -1 4 5 \...

What is the probability that a number selected at random from the numbers

Question: What is the probability that a number selected at random from the numbers 3, 4, 5, ....9 is a multiple of 4? Solution: GIVEN: numbers are 3, 4, 5, 69 TO FIND: Probability of Getting multiple of 4 Total number is 93+1=7 Numbers which are multiple of 4 between 3 and 9 are 4 and 8 Total number which are multiple of 4 between 3 to 9 is 4 and 8 is 2 We know that PROBABILITY = Hence probability of getting a number which is a multiple of 4 is Hence probability of getting a number which is a m...

A die is thrown once. What is the probability

Question: A die is thrown once. What is the probability of getting a number greater than 4? Solution: GIVEN: A dice is thrown once TO FIND: Probability of getting a number greater than 4 Total number on a dice is 6. Numbers greater than 4 are 5 and 6 Total number of numbers greater than 4 is 2 We know that PROBABILITY = Hence probability of getting a number greater than 4 is equal to Hence probability of getting a number greater than 4 is...

If w is a complex cube root of unity, show that

Question: Ifwis a complex cube root of unity, show that $\left(\left[\begin{array}{ccc}1 w w^{2} \\ w w^{2} 1 \\ w^{2} 1 w\end{array}\right]+\left[\begin{array}{ccc}w w^{2} 1 \\ w^{2} 1 w \\ w w^{2} 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ Solution: Here, $\mathrm{LHS}=\left(\left[\begin{array}{ccc}1 w w^{2} \\ w w^{2} 1 \\ w^{2} 1 w\end{array}\right]+\left[\begin{array}{ccc}w w^{2} 1 \\ w^{2} 1 w \\ w ...

Two coins are tossed simultaneously.

Question: Two coins are tossed simultaneously. Find the probability of getting exactly one head. Solution: GIVEN: Two coins are tossed simultaneously. TO FIND: Probability of getting exactly one head. When two coins are tossed then the outcome will be TT, HT, TH, HH. Hence total number of outcome is 4. Exactly one head we get 2 times Hence total number of favorable outcome i.e. exactly one head is 2 We know that PROBABILITY = Hence probability of getting exactly one head = Hence probability of g...