what is the probability that burglar is still unhurt?The probability that the burglar will be hit by a bullet is $0.6 .$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2,...
Answer : let A denote the event that the chosen individual is female and B denote the event that the chosen individual is over 50 years old. Given : Town consists of 6000 people, 1200 are over 50...
Answer : Let A denote the event that a patient visiting a denist will have a tooth extracted and B denote the event that a patient will have a cavity filled Given : P(A) = 0.06 , P(B) = 0.2 , P(A or...
Answer : Let A denote the event that a person will get electrification contract and B denote the event that the person will get a plumbing contract Given : P(A) = 2/5 , P(not B) =4/5 , P(A or B)...
Answer : let A denot the event that Hemant passes in english and B denote the event that hemant passes in hindi . Given : P(A) = 2/3, P(B) = 5/9 ,P(A and B) = 2/5 To find : Probability that he will...
Answer : Given: Math students = 30% Chemistry Students = 20% Math & Chemistry both = 10% To Find: P(Math or Chemistry) Now, P(Math) = 30/100= 0.30 P(Chemistry) = 20/100 = 0.20 P(Math ∩...
Answer : Given : two dice are tossed once To find : Probability of getting an even number on the first die or a total 8. The formula used : Probability = P(A or B) = P(A) + P(B) - P(A and B) A die...
Answer : Given : A die is thrown twice To find : Probability that at least one of the two throws comes up with the number 4 The formula used : Probability = A die is numbered from 1 to 6 When a die...
Answer : let A denote the event that the number is divisible by 4 and B denote the event that the number is divisible by 4. P(A or B) = P(A) + P(B) - P(A and B) To find : Probability that the number...
Answer : let A denote the event that the card drawn is spade and B denote the event that card drawn is king. In a pack of 52 cards, there are 13 spade cards and 4 king cards Given : P(A) = 13/52 ,...
that card drawn is the heart. In a pack of 52 cards, there are 4 queen cards and 13 heart cards Given : P(A) = 4/52 , P(B) =13/52 To find : Probability that card drawn is either a queen or heart =...
Answer : let A denote the event that the card drawn is king and B denote the event that card drawn is queen. In a pack of 52 cards, there are 4 king cards and 4 queen cards Given : P(A) = 4/52 ,...
Answer : let A denote the event that a company executive will travel by plane and B denote the event of him travelling by train Given : P(A) = 2/5, P(B) =1/3 To find : Probability of a company...
Answer : Given : A,B,C are mutually exclusive events and exhaustive events P(B) = (3/2) P(A) and P(C) = (1/2) P(B) To find : P(A) Formula used : P(A) + P(B) + P(C) = 1 For mutually exclusive events...
Answer : Given : A and B are mutually exclusive events P(not A) = P(B ) = 0.65 , P(A or B) = 0.65 To find : P(B) Formula used : P(A) = 1 – P() P(A or B) = P(A) + P(B) - P(A and B) For mutually...
Answer : Given : A and B are mutually exclusive events P(A) = 1/2, P(B) = 1/3 To find : P(A or B) Formula used : P(A or B) = P(A) + P(B) - P(A and B) For mutually exclusive events A and B, P(A and...
Answer : (i) Given : P(A) = 0.25, P(A or B) = 0.5 and P(B) = 0.4 To find : P(A and B) Formula used : P(A or B) = P(A) + P(B) - P(A and B) Substituting in the above formula we get, 0.5 = 0.25 + 0.4 –...
Answer : Given : P( A) = 0.4, P(A or B) = 0.7 and P(A and B) = 0.3 To find : P(B) Formula used : P(A) = 1 – P() P(A or B) = P(A) + P(B) - P(A and B) We have P(A ) = 0.4 P(A) = 1 – 0.4 = 0.6 We get...
Answer : Given : P(A) = 0.4, P(A or B) = 0.6 and P(B) = 0.5 To find : P(A and B) Formula used : P(A or B) = P(A) + P(B) - P(A and B) Substituting in the above formula we get, 0.6 = 0.4 + 0.5 – P(A...
Answer : Given : P(A) = 0.60, P(A or B) = 0.85 and P(A and B) = 0.42 To find : P(B) Formula used : P(A or B) = P(A) + P(B) - P(A and B) Substituting in the above formula we get, 0.85 = 0.60 + P(B) –...
Answer : Total no.of cases will be 6 x 6 x 6 = 216(because each die can have values from 1 to 6) Desired outcomes are those whose sum up to 5. Desired outcomes are (1, 1, 3), (1, 3, 1), ( 1, 2, 2),...
Answer : All numbers are different (given in question), this will be the same as picking r different objects from n objects which is ncr Here, n= 20 and r = 6(as we have to pick 6 different objects...
Answer : As repetition is not allowed total no.of cases possible is 9×8×7(because if one of the numbers occupies a wheel, then the other wheel cannot be occupied by this number, i.e. next wheel have...
Answer : Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3),...
Answer : (i) We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is Given, probability We know, probability a/a+b = 5/14 So, a = 5 and a+b...
Answer : We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is , similarly, if odds are not in the favor of the occurrence an event are...
Answer : We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is , which indirectly came from Probability of the occurrence of an event...
Answer : We know that, Probability of occurring = 1 - the probability of not occurring Given the probability of occurrence 7/10 Therefore, the probability of not occurrence = 1- 7/10 7/10...
Answer : (i) We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is Given, probability = 5/14 We know, probability a/a+b 5/14 . So, a = 5...
Answer : We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur similarly, if odds are not in the favor of the occurrence an event are a:b,...
Answer : We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to Where, Total no.of desired outcomes = a, and total no.of outcomes = a+b Given a = 8,...
Answer : We know that, 7/10 Therefore, the probability of not occurrence Probability of occurring = 1 - the probability of not occurring Given the probability of occurrence 1-7/10 Conclusion:...
Answer : We know that, Probability of occurring = 1 - the probability of not occurring Let’s calculate for the probability of not occurring, i.e. probability such that both of them don’t have a...
Answer : We know that, Probability of occurrence of an event So, we want another Tuesday that to from that 1 day left(as there is only one Tuesday left after 52 weeks)An ordinary year has 365 days...
Answer : We know that, Probability of occurrence of an event Desired output is picking a number which is multiple of 2 or 3. So, desire outputs are 2, 3, 4, 6, 8, 9, 10, 12. Total no.of desired...
Answer : We know that, Probability of occurrence of an event Desired output is a number greater than 3 and less than 10.Total no.of outcomes are 52 There will be four sets of each card naming A, 1,...
Answer : (i) We know that, Probability of occurrence of an event Total possible outcomes are alphabets from a to z Desired outcomes are a, e, i, o, u Conclusion: Probability of choosing a...
Answer : We know that, Probability of occurrence of an event (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),...
Answer : We know that, Probability of occurrence of an event Let T be tails and H be heads Total possible outcomes = TTT, TTH, THT, HTT, THH, HTH, HHT, HHH Desired outcomes are at least two heads or...
at least 2 tails Answer : We know that, Probability of occurrence of an eventLet T be tails and H be heads Total possible outcomes = TTT, TTH, THT, HTT, THH, HTH, HHT, HHH Desired outcomes are at...
Answer : We know that, Probability of occurrence of an event Total possible outcomes = TTT, TTH, THT, HTT, THH, HTH, HHT, HHHLet T be tails and H be heads Desired outcomes are at most two tails. So,...
Answer : We know that, Probability of occurrence of an event Total possible outcomes = TTT, TTH, THT, HTT, THH, HTH, HHT, HHH Desired outcomes are exactly one tail. So, desired outputs are THH, HTH,...
Answer : We know that, Probability of occurrence of an event Let T be tails and H be heads Total possible outcomes = TTT, TTH, THT, HTT, THH, HTH, HHT, HHH Desired outcomes are exactl two tails. So,...
Answer : We know that, Probability of occurrence of an event Let B be Boy and G be Girl Total possible outcomes are BB, BG, GB, GG Our desired outcome is at least one boy. So, BB, BG, GB are desired...
Answer : We know that, Probability of occurrence of an event Total no.of outcomes = 10+25 = 35 Desired outcomes are prizes. Total no.of desired outcomes = 10 Therefore, the probability of getting a...
Answer : We know that, Probability of occurrence of an event By permutation and combination, total no.of ways to pick r objects from given n objects is nCr Now, total no.of ways to pick a ball from...
Answer : We know that, Probability of occurrence of an event By permutation and combination, total no.of ways to pick r objects from given n objects is nCr Now, total no.of ways to pick a ball from...
Answer : We know that, Probability of occurrence of an event By permutation and combination, total no.of ways to pick r objects from given n objects is nCr Now, total no.of ways to pick a ball from...
Answer : We know that, Probability of occurrence of an event By permutation and combination, total no.of ways to pick r objects from given n objects is nCr Now, total no.of ways to pick a ball from...
Answer : We know that Probability of occurrence of an event By permutation and combination, total no.of ways to pick r objects from given n objects is nCr Now, total no.of ways to pick a ball from...
Answer : We know that, Probability of occurrence of an event By permutation and combination, total no.of ways to pick r objects from given n objects is nCr Now, total no.of ways to pick a ball from...
Answer : We know that, Probability of occurrence of an event (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),...
Answer : We know that, Probability of occurrence of an event Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3,...
Answer : We know that, Probability of occurrence of an event Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3,...
Answer : We know that, Probability of occurrence of an event (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),...
Answer : We know that, Probability of occurrence of an event (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),...
Answer : We know that, Probability of occurrence of an event (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),Total outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),...
Answer : (i) We know that, Probability of occurrence of an event (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),Outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),...
Answer : We know that, Probability of occurrence of an event As 4, 5 are two numbers between 3 and, so the desired outcomes are 3, 6, and total outcomes are 1, 2, 3, 4, 5, 6 3/6 Therefore, total...
Answer : We know that, Probability of occurrence of an event 2/6As 3, 6 are multiples up to 6, so the desired outcomes are 3, 6, and total outcomes are 1, 2, 3, 4, 5, 6 Therefore, total no.of...
Answer : We know that, Probability of occurrence of an event outcomes are 1, 2, 3, 4, 5, 6As 2, 3, 5 are prime numbers up to 6, so the desired outcomes are 2, 3, 5, and total 1/2 Conclusion:...
Answer : We know that, Probability of occurrence of an event outcomes are 1, 2, 3, 4, 5, 6As 1, 3, 5 are odd numbers up to 6, so the desired outcomes are 1, 3, 5, and total 2/3 Therefore, total...
Answer : We know that, Probability of occurrence of an event 2/3Total outcomes are 1, 2, 3, 4, 5, 6, and the desired outcomes are 2, 3 Therefore, total no.of outcomes are 6, and total no.of desired...
Answer : We know that, Probability of occurrence of an event Therefore, total no.of outcomes are 6, and total no.of desired outcomes are 1Total outcomes are 1, 2, 3, 4, 5, 6, and the desired outcome...
Answer : We know that Probability of occurrence of an event Hence the total no.of outcomes are 2 (i.e. heads and tails)Total outcomes of the coin are tails and heads And the desired output is tail....
Answer : The first three terms of a G.P. are:a,ar,ar2 The first six terms of a G.P. are:a,ar,ar2,ar3,ar4,ar5 It is given that the ratio of the sum of first three terms is to that of first six terms...
Answer : To prove: xb - c. yc - a. za - b = 1….(i) It is given that a,b,c are in A.P. ⇒ 2b = a + c…(ii) And x,y,z, are in G.P. ⇒ y2 = xz ⇒ x = y2/z Substitute this value of x in equation (i),we get...
Balls are drawn at random, So, the probability that at least one is white is, In a trial the probability of selecting a white ball is $\frac{5}{20}$ So, in 4 trials the probability that at least one...
(i) Balls are drawn at random, So, the probability that none is white is, In a trial the probability of selecting a non-white ball is $\frac{15}{20}$ So, in 4 trials it will be,...
Answer : It is given that: a1/x = b1/y = c1/z Let a1/x = b1/y = c1/z = k ⇒ a1/x = k ⇒ (a1/x)x = kx…(Taking power of x on both sides.) ⇒ a1/x × x = kx ⇒ a = kx Similarly b = ky And c = kz It is given...
, what is the probability that out of 10 men, now 60, at least 8 will live to be 
The probability that a man aged 60 will live to be 70 is $0.65$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots...
The probability that the bird will be shot, is $1 / 3$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots \ldots ....
Answer : We have been given that 2nd, 3rd and 6th terms of an AP are the three consecutive terms of a GP Let the three consecutive terms of the G.P. be a,ar,ar2. Where a is the first consecutive...
. What is the probability that he will knock down fewer than 2 hurdles?The probability that the hurdle will be cleared is $5 / 6$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots...
Answer : Let, tp + q = m = Arp + q - 1 = Arp - 1rq And tp - q = n = Arp - q - 1 = Arp - 1r - q We know that pth term = Arp - 1 ∴ m × n = A2r2p - 2 ⇒ Arp - 1 = (mn)1/2 ⇒ pth term = (mn)1/2 Ans: pth...
. If he fires 7 times, what is the probability of his hitting the target at least twice?Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2, \ldots \ldots . n \text { and } q=(1-p), n=7 \\ p= q=? \end{array}$...
of the operations performed by a certain doctor were successful. If the doctor performs 4 operations in a day, what is the probability that at least 3 operations will be successful?The probability that the operations performed are successful is $=0.8$ The probability that at least three operations are successful is $=\mathrm{P}(3)+\mathrm{P}(4)$ $\begin{array}{l} \Rightarrow{...
Answer : Given: 5th term of a GP is 2. To find: the product of its first nine terms. First term is denoted by a, the common ratio is denote by r. ∴ ar4 = 2 We have to find the value of: a × ar1 ×...
Answer :
Find the probability that all the cars reach the finishing line without any accident.The probability that any one of them has an accident is $0.1$. The probability any car reaches safely is $0.9 .$ The probability that all the cars reach the finishing line without any accident is...
The probability that the called number is busy is $\frac{1}{15}$ Using Bernoulli's Trial we have, $\begin{array}{l} P(\text { Success }=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)} \\ x=0,1,2,...
Answer : Let, x=2.134134134… …(i) Multiplying this equation by 1000 on both the sides so that repetitive terms cancel out and we get: 1000x=2134.134134134… …(ii) Equation (ii)-(i), ⇒...
Answer : Let, x=0.423423423… …(i) Multiplying this equation by 1000 on both the sides so that repetitive terms cancel out and we get: 1000x=423.423423423… …(ii) Equation (ii)-(i), ⇒...
defective items. A sample of 6 items is randomly chosen. Find the probability that this sample contains at least 3 defective itemsThe probability that the item is defective $=\frac{1}{10}=p$ The probability that the bulb will not fuse $=1-\frac{1}{10}=\frac{9}{10}=q$ Using Bernoulli's we have, $\begin{array}{l} P(\text {...
Answer : Let, x=0.125125125… …(i) Multiplying this equation by 1000 on both the sides so that repetitive terms cancel out and we get: 1000x=125.125125125… …(ii) Equation (ii)-(i), ⇒...
defective items. A sample of 6 items is randomly chosen. Find the probability that this sample contains.(i) The probability that the item is defective $=\frac{1}{10}=p$ The probability that the bulb will not fuse $=1-\frac{1}{10}=\frac{9}{10}=\mathrm{q}$ Using Bernoulli's we have, $\begin{array}{l}...
find the probability that out of 5 such bulbs not more than one will fuse after 6 months of useThe probability that the bulb will fuse $=0.05=p$ The probability that the bulb will not fuse $=1-0.05=0.95=q$ Using Bernoulli's we have, $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$...
find the probability that out of 5 such bulbs(i) The probability that the bulb will fuse $=0.05=\mathrm{p}$ The probability that the bulb will not fuse $=1-0.05=0.95=q$ Using Bernoulli's we have, $\begin{array}{l} P(\text { Success }=x)={...
(i) Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n \text { and } q=(1-p), n=5$ The probability of success, i.e. the bulb is...
defective items in a large bulk of times. Find the probability that a sample of 8 items will include not more than one detective item.Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . .$ and $q=(1-p), n=8$ The probability of success, i.e. the bulb is defective...
is considered a success, find the probability of getting(i) Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots \ldots$ and $q=(1-p), n=7$ the favourable outcomes,...
is considered a success, find the probability of getting(i) Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and $q=(1-p), n=7$ the favourable outcomes,...
As the pair of die is thrown 4 times, The total number of outcomes $=36$ Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and...
The total outcomes $=36$ The favourable outcomes are $(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),(4,1),(4,2),(4,3),(4,5),(4,6)$ Thus, the probability $=$ favourable outcomes/total outcomes $\Rightarrow...
Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and $q=(1-p)$ We know that the favourable outcomes of getting at most 2 successes...
(i) Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots \ldots \text { and } q=(1-p)$ We know that the favourable outcomes of getting exactly...
Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and $q=(1-p)$ As the die is thrown 6 times the total number of outcomes will be...
Answer : To find: The quadratic equation. Given: (i) AM of roots of quadratic equation is 10 (ii) GM of roots of quadratic equation is 8 Formula used: (i) Arithmetic mean between (ii) Geometric mean...
Answer : To prove: Product of n geometric means between a and b is equal to the nth power of the single GM between a and b. Formula used:(i) Geometric mean between (ii) Sum of n terms of A.P. Let...
Answer : To prove: b2 is the AM between x2 and y2. Given: (i) a, b, c are in AP x is the GM between a and b y is the GM between b and c Formula used: (i) Arithmetic mean between (ii) Geometric mean...
Answer : To find: Four geometric Mean Given: The numbers 6 and 192 Formula used: (i) r where n is the number of geometric mean Let G1, G2, G3 and G4 be the three geometric mean Then r ⇒ r = 2 G1 =...
Answer : To find: Three geometric Mean ???? and 432 Given: The numbers ????/3 and 432 Formula used: where n is the number of geometric mean Let G1, G2 and G3 be the three geometric mean
Answer : To find: Two geometric Mean Given: The numbers are 9 and 243 geometric mean Let G1 and G2 be the three geometric mean Then r ⇒ r = 3 G1 = ar = 9×3 = 27 G2 = ar2= 9×32 = 9×9 = 81 Two...
Answer : (i) 5 and 125 To find: Geometric Mean Given: The numbers are 5 and 125 Formula used: (i) Geometric mean between Geometric mean of two numbers = 25 The geometric mean between 5 and 125 is 25...
Answer : (i) AM = 25 and GM = 20 To find: Two positive numbers a and b Given: AM = 25 and GM = 20 Formula used: (i) Arithmetic mean between (ii) Geometric mean between Arithmetic mean of two...
Answer : To prove: x2, b2, y2 are in AP. Given: a, b, c are in AP, and a, x, b an b, y, c are in GP Proof: As, a,b,c are in AP ⇒ 2b = a + c … (i) As, a,x,b are in GP ⇒ x2 = ab … (ii) As, b,y,c are...
Answer : To prove: a, (a – b) and (d – c) are in GP. Given: a, b, c are in AP, and a, b, d are in GP Proof: As a,b,d are in GP then b2 = ad … (i) As a, b, c are in AP 2b = (a + c) … (ii) Considering...
Answer : To prove: p, q, r are in GP Given: (p2 + q2), (pq + qr), (q2 + r2) are in GP Formula used: When a,b,c are in GP, b2 = ac Proof: When (p2 + q2), (pq + qr), (q2 + r2) are in GP, (pq + qr)2 =...
Answer : To prove: (a2 – b2), (b2 – c2), (c2 – d2) are in GP. Given: a, b, c are in GP From the above, we can have the following conclusion Formula used: When a,b,c are in GP, b2 = ac Proof: When...
(i) Using Bernoulli's Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . . n$ and $q=(1-p)$ As the die is thrown 6 times the total number of outcomes will...
Answer : To prove: (a2 + b2), (ab + bc), (b2 + c2) are in GP Given: a, b, c are in GP Formula used: When a,b,c are in GP, b2 = ac Proof: When a,b,c are in GP, b2 = ac … (i) Considering (a2 + b2),...
As 10 coins are tossed simultaneously the total number of outcomes are $2^{10}=1024$. the favourable outcomes of getting at least 4 heads will be ${ }^{10} \mathrm{C}_{4}+{ }^{10} \mathrm{C}_{5}+{...
Answer : To prove: a3, b3, c3 are in GP Given: a, b, c are in GP Proof: As a, b, c are in GP ⇒ b2 = ac Cubing both sides = common ratio = r From the above equation, we can say that a3, b3, c3 are in...
(i) As 10 coins are tossed simultaneously the total number of outcomes are $2^{10}=1024$. the favourable outcomes of getting exactly 3 heads will be ${ }^{10} \mathrm{C}_{3}=120$ Thus, the...
Answer : To prove: a2, b2, c2 are in GP Given: a, b, c are in GP Proof: As a, b, c are in GP ⇒ b2 = ac … (i) Considering b2, c2 = common ratio = r ⇒ [From eqn. (i)] ⇒ = r Considering a2,...
As the coin is tossed 6 times the total number of outcomes will be $2^{6}=64$ And we know that the favourable outcomes of getting at most 4 heads will be ${ }^{6} \mathrm{C}_{0}+{ }^{6}...
(i) As the coin is tossed 6 times the total number of outcomes will be $2^{6}=64$ And we know that the favourable outcomes of getting exactly 4 heads will be ${ }^{6} c_{4}=15$ Thus, the probability...
As 7 coins are tossed simultaneously the total number of outcomes are $2^{7}=128$. The favourable number of outcomes that a tail appears an odd number of times will be, ${ }^{7} \mathrm{C}_{1}+{...
As the coin is tossed 5 times the total number of outcomes will be $2^{5}=32$. And we know that the favourable outcomes of a head appearing even number of times will be, That either the head appears...
Hence Proved (iii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2 To prove: (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2 Given: a, b, c, d are in GP Proof: When a,b,c,d are in GP thenFrom...
As the coin is tossed 6 times the total number of outcomes will be $2^{6}$. And we know that the favourable outcomes of getting at least 3 heads will be ${ }^{6} c_{3}+{ }^{6} c_{4}+{ }^{6} c_{5}+{...
![\mathrm{A}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b68ab71b95db056eec8874de20c73761_l3.png "Rendered by QuickLaTeX.com")
, prove that ![\mathrm{A}^{\mathrm{n}}=\left[\begin{array}{ll}1 & \mathrm{n} \\ 0 & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c75542afc0bd2c36924570507619d8df_l3.png "Rendered by QuickLaTeX.com")
for all 
.Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{1} & \mathbf{1} \\ \mathbf{0} & \mathbf{1}\end{array}\right)$. We need to show: $A^{n}=\left(\begin{array}{ll}1 & n \\ 0...
![\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]=\mathrm{O}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-95fdea3519bebde2f1b28d9c4f95bf38_l3.png "Rendered by QuickLaTeX.com")
, find 
Solution: We have $A=\left(\begin{array}{ccc}2 & 1 & 2 \\ 1 & 0 & 2 \\ 0 & 2 & -4\end{array}\right), B=\left(\begin{array}{lll}x & 4 & 1\end{array}\right)$ and...
Answer : To prove: (a + b + c)2 = 3(ab + bc + ca). Given: (a – b), (b – c), (c – a) are in GP Formula used: When a,b,c are in GP, b2 = ac As, (a – b), (b – c), (c – a) are in GP ⇒ (b – c)2 = (a – b)...
Answer : To prove: Given: a, b, c are in GP Formula used: When a,b,c are in GP, b2 = ac a, b, c are in GP, ⇒ b2 = ac … (i) Taking LHS Substituting the value b2 = ac from eqn. (i) Substituting the...
Answer : To find: Three numbers Given: Three numbers are in G.P. Their sum is 56 Formula used: When a,b,c are in GP, b2 = ac Let the three numbers in GP be a, ar, ar2 According to condition :- a +...
Answer : To find: Three numbers Given: Three numbers are in A.P. Their sum is 21 Formula used: When a,b,c are in GP, b2 = ac Let the numbers be a - d, a, a + d According to first condition a + d + a...
Answer : To find: The numbers Given: Three numbers are in A.P. Their sum is 15 Formula used: When a,b,c are in GP, b2 = ac Let the numbers be a - d, a, a + d According to first condition a + d + a...
Answer : To find: Value of k Given: k + 12, k – 6 and 3 are in GP Formula used: (i) when a,b,c are in GP b2 = ac As, k + 12, k – 6 and 3 are in GP ⇒ (k – 6)2 = (k + 12) (3) ⇒ k2 – 12k + 36 = 3k + 36...
Answer : To prove: log an, log bn, log cn are in AP. Given: a, b, c are in GP Formula used: (i) log ab = log a + log b As a, b, c are in GP ⇒ b2 = ac Taking power n on both sides ⇒ b2n = (ac)n...
To prove: pth, qth and rth terms of any GP are in GP. Given: (i) p, q and r are in AP The formula used: (i) General term of GP, As p, q, r are in A.P.
Answer : To find: The amount after five years Given: (i) Principal – 156250 Time – 5 years Rate – 20% per annum
Answer : To find: Present population of the village Given: (i) Three years back population - 10000 Time – 3 years Rate – 20% per annum Number of people migrated on the very first year is 20% of...
Answer : To find: The amount after three days Given: (i) Principal – 80000 Time – 3 days = 98.63 Rate – 15% per annum Deduction = P × R × T The final amount after deduction = 80000 – 98.63 =...
Answer : To find: The amount after three years Given: (i) Principal – 15625 Time – 3 years Rate – 8% per annum Formula used: A = 19683 Ans) 19683
Answer : ‘Tn’ represents the nth term of a G.P. series. Tn = arn-1 ⇒486 = a(3)n-1 ⇒486 = a( 3n ÷ 3) ) ⇒486 × 3 = a(3n) ⇒1458 = a(3n ) ………(i) Sum of a G.P. series is represented by the formula, when...
Sum of a G.P. series is represented by the formula, when r>1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’...
Answer : Tn represents the nth term of a G.P. series. r = 6 ÷ 3 = 2 Tn = arn-1 ⇒1536 = 3 × 2n-1 ⇒1536 ÷ 3 = 2n ÷ 2 ⇒1536 ÷ 3 × 2 = 2n ⇒1024 = 2n ⇒210 = 2n ∴ n = 10 Sum of a G.P. series is...
Answer : Sum of a G.P. series is represented by the formula, when r≠1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and...
Answer : In this question, we will try to rewrite the given sum of the progression like the formula for the sum a G.P. series. It is given that Sn = ( 2n – 1) The formula for the sum of a G.P....
Answer : The given expression can be written as = (x2+ xy) + (x4 + x2y2 ) + (x 6 + x3y3 ) + …. To n terms = (x2 + x4 + x6 + … to n terms ) + ( xy + x2y2 + x3y3 + … to n terms )
Answer : Sum of a G.P. series is represented by the formula, when r>1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio...
Answer : Sum of a G.P. series is represented by the formula, , when r>1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’...
Answer : Sum of a G.P. series is represented by the formula, when |r|<1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio...
Answer : Sum of a G.P. series is represented by the formula, when |r|<1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio...
Answer : Sum of a G.P. series is represented by the formula, when |r|<1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio...
Answer : Given data is, x2 – 3x + p = 0 → (1) a and b are roots of (1) So, (x + a)(x + b) = 0 x2 - (a + b)x + ab = 0 So, a + b = 3 and ab = p → (2) Given data is, x2 – 12x + q = 0 → (3) c and d are...
Answer : Cross multiplying (1) and expanding, (a + bx)(b – cx) = (b + cx)(a-bx) ab – acx + b2x – bcx2 = ba –b2x + acx – bcx2 2b2x = 2acx b2 = ac → (i) If three terms are in GP, then the middle term...
Answer : We need to prove that the product of the terms equidistant from the beginning and end is the product of first and last terms in a finite GP. Let us first consider a finite GP. A, AR,...
Answer : As per the question, a, b and c are the pth, qth and rth term of GP. Let us assume the required GP as A, AR, AR2, AR3… Now, the nth term in the GP, an = ARn-1 pth term, ap = ARp-1 = a → (1)...
The common ratio, r = 3 The last term in the given GP is an = 162. Second last term in the GP = an-1 = arn-2 Starting from the end, the series forms another GP in the form, arn-1, arn-2, arn-3….ar3,...
Answer : The given GP is 8, 4, 2… 1/1024 First term in the GP, a1 = a = 8 Second term in the GP, a2 = ar = 4 The common ratio, The last term in the given GP is 1/1024 Second last term in the GP =...
Answer : It is given that the first term of GP is -3. So, a = -3 It is also given that the square of the second term is equal to its 4th term. ∴ (a2)2 = a4 nth term of GP, an = arn-1 So, a2 = ar; a4...
Answer : The nth term of a GP is an = arn-1 It’s given in the question that 5th term of the GP is 80 and 8th term of GP is 640. So, a5 = ar4 = 80 → (1) a8 = ar7 = 640 → (2) Common ratio, r = 2, ar4...
Answer : Given GP is √3, 3, 3√3…. The given GP is of the form, a, ar, ar2, ar3…. Where r is the common ratio. First term in the given GP, a1 = a = √3 Second term in GP, a2 = 3 Now, the common ratio,...
Answer : Given GP is 1 , −1 , 1. . .. The given GP is of the form, a, ar, ar2, ar3…. Where r is the common ratio. Let us consider -128as the nth term of the GP. Now, nth term of GP is, an =...
Answer : Given GP is 3, 6, 12, 24…. The given GP is of the form, a, ar, ar2, ar3…. Where r is the common ratio. First term in the given GP, a1 = a = 3 Second term in GP, a2 = 6 Let us consider 3072...
Answer : Given GP is . The given GP is of the form, a, ar, ar2, ar3…. Where r is the common ratio. Now, nth term of GP is, an = arn – 1 So,...
Answer : Given GP is 0.4, 0.8, 1.6…. The given GP is of the form, a, ar, ar2, ar3…. Where r is the common ratio. First term in the given GP, a1 = a = 0.4 Second term in GP, a2 = 0.8 Now, nth term of...
Answer : Given GP is 2, 2√2, 4, 8√2 ….. The given GP is of the form, a, ar, ar2, ar3…. Where r is the common ratio. First term in the given GP, a1 = a = 2 Second term in GP, a2 = 2√2 Now, nth term...
Answer : Given: GP is 2, 6, 18, 54…. The given GP is of the form, a, ar, ar2, ar3…. Where r is the common ratio. Now, nth term of GP is, an = arn – 1 First term in the given GP, a1 = a = 2 Second...
![A=\left[\begin{array}{rr}1 & -1 \\ 2 & -1\end{array}\right], B=\left[\begin{array}{rr}a & -1 \\ b & -1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7966787aabbac62e99a047018387f869_l3.png "Rendered by QuickLaTeX.com")
and 
then find the values of 
and 
.Solution: We have $A=\left(\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right), B=\left(\begin{array}{ll}a & -1 \\ b & -1\end{array}\right)$ and $(A+B)^{2}=\left(A^{2}+\right.$...
![\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right] \cdot \mathrm{A}=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-633e2dddcd4fd5ca06e3048dd98f602c_l3.png "Rendered by QuickLaTeX.com")
.Solution: We have $\boldsymbol{B}=\left(\begin{array}{cc}\mathbf{5} & -\mathbf{7} \\ -\mathbf{2} & \mathbf{3}\end{array}\right)$ and $\boldsymbol{C}=\left(\begin{array}{cc}-\mathbf{1 6}...
and 
, when ![\left[\begin{array}{cc} 3 & -4 \\ 1 & 2 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \end{array}\right]=\left[\begin{array}{l} 3 \\ 11 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-16dd3658e32c62682e7e1c991959454b_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{cc}3 & -4 \\ 1 & 2\end{array}\right), B=\left(\begin{array}{c}3 \\ 11\end{array}\right)$ and $X=\left(\begin{array}{l}x \\ y\end{array}\right)$. We...
![A=\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8fe923545e5bdad93e0724bc71eca803_l3.png "Rendered by QuickLaTeX.com")
and 
, find 
.Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{1} & \mathbf{2} \\ \mathbf{4} & \mathbf{- 3}\end{array}\right)$. Now addition/subtraction of two matrices is possible if...
![A=\left[\begin{array}{cc}a b & b^{2} \\ -a^{2} & -a b\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-69596ec6bdf3b660c44b6d9df9581f38_l3.png "Rendered by QuickLaTeX.com")
, show that 
.Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\boldsymbol{a} \boldsymbol{b} & \boldsymbol{b}^{2} \\ -\boldsymbol{a}^{2} & -\boldsymbol{a} \boldsymbol{b}\end{array}\right) .$ To...
![A=\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right], B=\left[\begin{array}{ccc}0 & 5 & -4 \\ -2 & 1 & 3 \\ 1 & 0 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a00e039f4079110b22d247fddf0823b8_l3.png "Rendered by QuickLaTeX.com")
and ![C=\left[\begin{array}{ccc}1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right] ;](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dda16b559739c22d0dee0f77fa7c40b0_l3.png "Rendered by QuickLaTeX.com")
verify that 
Solution: We have $A=\left(\begin{array}{ccc}1 & 0 & 2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right), B=\left(\begin{array}{ccc}0 & 5 & -4 \\ -2 & 1 & 3 \\ 1...
, when ![\mathrm{A}=\left[\begin{array}{cc} 2 & 3 \\ -1 & 4 \\ 0 & 1 \end{array}\right], \mathrm{B}=\left[\begin{array}{cc} 5 & -3 \\ 2 & 1 \end{array}\right] \text { and } \mathrm{C}=\left[\begin{array}{cc} -1 & 2 \\ 3 & 4 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fd66ff615b29326ba3d097b4946cee2e_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{cc}2 & 3 \\ -1 & 4 \\ 0 & 1\end{array}\right), B=\left(\begin{array}{cc}5 & -3 \\ 2 & 1\end{array}\right)$. and...