A fair coin is tossed three times. Let A, B and C be defined as follows: -Maths 9th

A fair coin is tossed three times. Let A, B and C be defined as follows: -Maths 9th
by

1 Answer

Answer :

The sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HHH, HHT, HTH, HTT}, B = {HHH, HHT, THH, THT} and C = {HHT, THH} Also, A ∩ B = {HHH, HHT}, B ∩ C = {HHT, THH}, C ∩ A = {HHT}P (A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{8}\) = \(rac{1}{2}\),  P(B) = \(rac{n(B)}{n(S)}\) = \(rac{4}{8}\) = \(rac{1}{2}\),  P(C) = \(rac{n(C)}{n(S)}\) = \(rac{2}{8}\) = \(rac{1}{4}\),P (A ∩ B) = \(rac{n[(A\,\cap\,B)]}{n(S)}\) = \(rac{2}{8}\) = \(rac{1}{4}\). Similarly, P (B ∩ C) = \(rac{2}{8}\) = \(rac{1}{4}\) and P (C ∩ A) = \(rac{1}{8}\).Now, P (A) . P (B) = \(rac{1}{2}\) x \(rac{1}{2}\) = \(rac{1}{4}\) = P (A ∩ B), therefore, A and B are independent events.P (B) . P (C) = \(rac{1}{2}\) x \(rac{1}{4}\) x \(rac{1}{8}\) ≠ P (B ∩ C) which is \(rac{1}{4}\) . Therefore, B and C are not independent events.P (C) . P (A) = \(rac{1}{4}\) x \(rac{1}{2}\) x \(rac{1}{8}\) = P(C ∩ A) Therefore, C and A are independent events.Three events A, B and C are independent if: (i) P (A ∩ B) = P (A). P (B), P (A ∩ C) = P (A). P (C) and P (B ∩ C) = P (B). P (C), i.e., if the events are pairwise independent and (ii) P (A ∩ B ∩ C) = P (A) . P (B) . P (C)
by

Related questions

Let a pair of fair coins be tossed. Here S = {HH, HT, TH, TT}. Consider the events A = {heads on the first coin} = {HH, HT}, -Maths 9th

Description : Let a pair of fair coins be tossed. Here S = {HH, HT, TH, TT}. Consider the events A = {heads on the first coin} = {HH, HT}, -Maths 9th

Last Answer : ThenP (A) = P (B) = P (C) = \(rac{2}{4}\) = \(rac{1}{2}\) andP (A ∩ B) = P ({HH}) = \(rac{1}{4}\), P (A ∩ C) = P ({HT}) = \(rac{1}{4}\)P ( ... C)Thus condition (i) is satisfied, i.e., the events are pairwise independent. But condition (ii) is not satisfied and so the three events are not independent

A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails. -Maths 9th

Description : Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails. -Maths 9th

Last Answer : (d) \(rac{1}{2}\)Let S be the sample space. Then, S = {HHH, HHT, HTH, HTT, THH, THT,TTH, TTT} ⇒ n(S) = 8 Let A : Event of getting more heads than number of tails. Then, A = {HHH, HHT, HTH, THH} ⇒ n(A) = 4∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{8}\) = \(rac{1}{2}.\)

A coin is tossed thrice and all eight outcomes are assumed equally likely. Find whether the events E -Maths 9th

Description : A coin is tossed thrice and all eight outcomes are assumed equally likely. Find whether the events E -Maths 9th

Last Answer : When a coin is tossed three times, the sample space is given by S = [HHH, HHT, HTH, THT, THH, HTT, TTH, TTT] E = {HHH, HTT, THT, TTH}, F = {TTT, HTH, THH, HHT}E ∩ F = ϕP(E) = \(rac{4}{8}\) = \(rac{1}{2}\ ... rac{1}{2}\) x \(rac{1}{2}\) x \(rac{1}{4}\) ≠ P(E ∩ F) ∴ E and F are not independent events.

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Last Answer : Total number of chances = 23 + 72 + 77 + 28 = 200 Number of chances of coming 2 heads = 72 therefore P( coming 2 heads)= 514 / 642 = 9 / 25

Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. -Maths 9th

Description : Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. -Maths 9th

Last Answer : Total number of outcomes = 360 Let the number of times ‘No Tail’ appeared be x Then, number of times ‘2 Tails’ appeared =3x Number of times ‘1 Tail’ appeared =2x Now, x + 2x + 3x =360 ⇒ 6x =360 ⇒ x= 60 P(of getting two tails)=(3 x 60) / 360 =1 / 2

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Last Answer : It is given that coin is tossed 200 times Total number of trials = 200 Number of events for getting less than three tails = 68 + 82 + 30 = 180 Probability of getting less than 3 tails =180 / 200 =9 / 10

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes ; -Maths 9th

Last Answer : Total number of chances = 23 + 72 + 77 + 28 = 200 Number of chances of coming 2 heads = 72 therefore P( coming 2 heads)= 514 / 642 = 9 / 25

Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. -Maths 9th

Description : Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. -Maths 9th

Last Answer : Total number of outcomes = 360 Let the number of times ‘No Tail’ appeared be x Then, number of times ‘2 Tails’ appeared =3x Number of times ‘1 Tail’ appeared =2x Now, x + 2x + 3x =360 ⇒ 6x =360 ⇒ x= 60 P(of getting two tails)=(3 x 60) / 360 =1 / 2

Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Description : Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes -Maths 9th

Last Answer : It is given that coin is tossed 200 times Total number of trials = 200 Number of events for getting less than three tails = 68 + 82 + 30 = 180 Probability of getting less than 3 tails =180 / 200 =9 / 10

Three coins were tossed 30 times simultaneously. -Maths 9th

Description : Three coins were tossed 30 times simultaneously. -Maths 9th

Last Answer : Frequency disribution of above data in tabular form is given as:

Two coins are tossed 1000 times and the outcomes are recorded as below : -Maths 9th

Description : Two coins are tossed 1000 times and the outcomes are recorded as below : -Maths 9th

Last Answer : Required probability = P(0 heads) + P(1 head) = 250/1000 + 550 / 1000 = 800/ 1000 =4 / 5 =0.8

Two coins are tossed 1000 times and the outcomes are recorded as below : -Maths 9th

Description : Two coins are tossed 1000 times and the outcomes are recorded as below : -Maths 9th

Last Answer : Required probability = P(0 heads) + P(1 head) = 250/1000 + 550 / 1000 = 800/ 1000 =4 / 5 =0.8

Two coins are tossed 1000 times and the outcomes are recorded as below: -Maths 9th

Description : Two coins are tossed 1000 times and the outcomes are recorded as below: -Maths 9th

Last Answer : P (at most one head) = P (0 head) + P (1 head) = 250/1000 + 550/1000 = 800/1000 = 4/5

Two coins are tossed simultaneously 500 times. -Maths 9th

Description : Two coins are tossed simultaneously 500 times. -Maths 9th

Last Answer : Since, frequency of one or more than one head = 100 + 270 = 370 Therefore, P (one or more heads) = 370/500 = 37/50

Three coins are tossed simultaneously -Maths 9th

Description : Three coins are tossed simultaneously -Maths 9th

Last Answer : Frequency of more than one tail = 135 + 85 = 220 ∴ P (more than one tail) = 220/500 = 11/25

A coin is tossed 500 times. Head occurs 343 times and tail occurs 157 times. Find the probability of each event.

Description : A coin is tossed 500 times. Head occurs 343 times and tail occurs 157 times. Find the probability of each event.

Last Answer : A coin is tossed 500 times. Head occurs 343 times and tail occurs 157 times. Find the probability of each event.

A coin is tossed 20 times and head occurred 12 times. How many times did tail occur?

Description : A coin is tossed 20 times and head occurred 12 times. How many times did tail occur?

Last Answer : A coin is tossed 20 times and head occurred 12 times. How many times did tail occur?

A single coin is tossed 7 times. What is the probability of getting at least one tail? a) 127/128 b) 128/127 c) 2/128 d) 4/128

Description : A single coin is tossed 7 times. What is the probability of getting at least one tail? a) 127/128 b) 128/127 c) 2/128 d) 4/128

Last Answer : Answer: A) Consider solving this using complement. Probability of getting no tail = P(all heads) = 1/128 P(at least one tail) = 1 – P(all heads) = 1 – 1/128 = 127/128

If two coins are tossed once, what is the probability of getting at least one head ? -Maths 9th

Description : If two coins are tossed once, what is the probability of getting at least one head ? -Maths 9th

Last Answer : When two coins are tossed once, there are four possible outcomes, i.e., S = {HH, HT, TH, TT} ∴ Total number of outcomes = n(S) = 4 Let A : Event of getting at least one head ⇒ A = {HH, HT, TH} ⇒ n(A) = 3∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{3}{4}.\)

Let R be a relation from A = {1, 2, 3, 4, 5, 6} to B = {1, 3, 5} which is defined as “x is less than y”. -Maths 9th

Description : Let R be a relation from A = {1, 2, 3, 4, 5, 6} to B = {1, 3, 5} which is defined as “x is less than y”. -Maths 9th

Last Answer : R = {a, b : a < b, a ∈ A, b ∈ B}, where A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 5}. ∴ R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} Domain of R = {1, 2, 3, 4} Range of R = {3, 5} Codomain of R = {1, 3, 5}.

Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Description : Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Last Answer : (d) An equivalence relation.Every triangle is similar to itself, so (T1, T1) ∈ R ⇒ R is reflexive. (T1, T2) ∈ R ⇒ T1 ~ T2 ⇒T2 ~ T1, ⇒ (T2, T1) ∈ R ⇒ R is symmetrictransitive. ∴ R is an equivalence relation.

Let R be a relation defined as a Rb if | a – b | > 0, then the relation is -Maths 9th

Description : Let R be a relation defined as a Rb if | a – b | > 0, then the relation is -Maths 9th

Last Answer : (d) Symmetric and transitive| a - a | = | 0 | = 0 so (a, a) ∉R ⇒ R is not reflexive(a, b) ∈ R ⇒ | a - b | > 0 ⇒ | b - a | > 0 ⇒ (b, a) ∈R (∵ | a - b | = | b - a |) ⇒ R is symmetric (a, b) ∈ R ... a, b, c. ∴ | a - b | > 0 and | b - c | > 0 ⇒ | a - c | > 0 ⇒ (a, c) ∈ R ⇒ R is transitive.

Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Description : Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Last Answer : (a) ReflexiveGiven, {(\(x\), y) : 2\(x\) – y = 10} Reflexive, \(x\) R \(x\) = 2\(x\) – \(x\) = 10 ⇒ \(x\) = 10 ⇒ y = 10 ∴ Point (10, 10) ∈ N ⇒ R is reflexive.

Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Description : Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Last Answer : (d) R = {(a, b) : b/a is odd} Since (2, 6) ∈R, the relation a and b are odd does not exist. Since (1, 5) and (1, 7) ∈R, the relation a and b are even does not exist. None of the ... \(rac{7}{1}\) = 7, \(rac{6}{2}\) = 3, quotients being all odd numbers, the relation b/a is odd exists.

Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Description : Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Last Answer : (c) {(2, 3), (5, 3), (5, 6), (11, 3), (11, 6), (11, 10)}

In tossing a coin 100 times head appears 56 times. -Maths 9th

Description : In tossing a coin 100 times head appears 56 times. -Maths 9th

Last Answer : P (head) = 56/100 = 0.56.

When an unbiased coin is tossed, the probability of getting a head is ______.

Description : When an unbiased coin is tossed, the probability of getting a head is ______.

Last Answer : When an unbiased coin is tossed, the probability of getting a head is ______.

one rupee coin is tossed twice. What is the probability of getting two consecutive heads ? A)1/2 B)1/4 C)3/4 D)4/3

Description : one rupee coin is tossed twice. What is the probability of getting two consecutive heads ? A)1/2 B)1/4 C)3/4 D)4/3

Last Answer : Answer: B) Probability of getting a head in one toss = 1/2 The coin is tossed twice. So 1/2 * 1/2 = 1/4 is the answer. Here's the verification of the above answer with the help of sample ... (H,H) whose occurrence is only once out of four possible outcomes and hence, our answer is 1/4.

A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is -Maths 9th

Description : A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is -Maths 9th

Last Answer : (c) \(rac{84 imes5^6}{6^{10}}\)In the first nine throws we should have three sixes and six non-sixes and a six in the tenth throw and thereafter whatever face appears, it doesn't matter. ∴ Required probability = 9C3 \(\bigg(rac{1} ... x 1 x 1 ............x 1 {10 times} = \(rac{84 imes5^6}{6^{10}}\).

As the number of tosses of a coin increases, -Maths 9th

Description : As the number of tosses of a coin increases, -Maths 9th

Last Answer : No. As the number of tosses of a coin increases, the ratio of the number of heads to the total number of tosses will be near to 1/2, not exactly 1/2.

A and B throw a coin alternately till one of them gets a ‘head’ and wins the game. Find their respective probabilities of winning . -Maths 9th

Description : A and B throw a coin alternately till one of them gets a ‘head’ and wins the game. Find their respective probabilities of winning . -Maths 9th

Last Answer : Let A : Event of A getting a head ⇒ \(\bar{A}\) : Event of A not getting a head ∴ P(A) = \(rac{1}{2}\) and P(\(\bar{A}\)) = 1 - \(rac{1}{2}\) = \(rac{1}{2}\)Similarly, B : Event of B ... exclusive events, as either of them will win, P(B winning the game first) = 1 - \(rac{2}{3}\) = \(rac{1}{3}\).

A coin and six faced die, both unbiased are thrown simultaneously. -Maths 9th

Description : A coin and six faced die, both unbiased are thrown simultaneously. -Maths 9th

Last Answer : (c) \(rac{1}{4}\)Let A : Event of getting a tail on the coin B : Event of getting an even number on the die. Then, P(A) = \(rac{1}{2}\)P(B) = \(rac{3}{6}\) = \(rac{1}{2}\) as B = {2,4,6}A and B being independent events ... die)= P(A ∩ B) = P(A) P(B) = \(rac{1}{2}\)x\(rac{1}{2}\) = \(rac{1}{4}\).

Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Description : Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Last Answer : (i) Frequency distribution table (ii) From the above frequency distribution table, we observe that number of children in the class - interval 15 - 20 is 2. So, 2 children view television for 15 hours or more than 15 hours a week .

Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Description : Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Last Answer : (i) Frequency distribution table (ii) From the above frequency distribution table, we observe that number of children in the class - interval 15 - 20 is 2. So, 2 children view television for 15 hours or more than 15 hours a week .

The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Description : The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Last Answer : NEED ANSWER

The frequency distribution has been represented graphically as follows. -Maths 9th

Description : The frequency distribution has been represented graphically as follows. -Maths 9th

Last Answer : NEED ANSWER

The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Description : The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Last Answer : (b) Since, the difference between mid value is 5. So, the corresponding class to the class mark 20 must have difference 5. Therefore, option (c) and (d) are wrong. Since, the mid value is 20 which can get only, if we take option (b)

The frequency distribution has been represented graphically as follows. -Maths 9th

Description : The frequency distribution has been represented graphically as follows. -Maths 9th

Last Answer : No, here the widths of the rectangles are varying, so we need to make certain modifications in the length of the rectangles so that the areas are proportional to the frequencies. We proceed as ... Now, we get the following modified table So, the correct histogram with varying width is given below

A wooden bookshelf has external dimensions as follows: -Maths 9th

Description : A wooden bookshelf has external dimensions as follows: -Maths 9th

Last Answer : Area of the bookshelf to be polished = Area of the five complete surfaces + Area of 2 rectangles of dimensions 110 cm x 5 cm in the front + Area of 4 rectangles of dimensions 75 cm x 5 cm in the ... of painting = ₹ 10/100 x 19350 = ₹1935 Hence, total expenses required = ₹(4340 + 1935) = ₹6275

The points scored by a basketball team in a series of matches are as follows: -Maths 9th

Description : The points scored by a basketball team in a series of matches are as follows: -Maths 9th

Last Answer : Arranging the data in ascending order, we get 2, 5, 7, 7, 8, 10, 10, 10, 14, 17, 18, 24, 25, 27, 28, 48 Since number of observations (n) = 16, which is even. Therefore, median is the mean of (n/2) ... ∴ Median = 10 + 14/2 = 12 As 10 occurs most frequently, i.e., three times. So, the mode is 10.

The blood groups of 30 students of class VIII are recorded as follows: -Maths 9th

Description : The blood groups of 30 students of class VIII are recorded as follows: -Maths 9th

Last Answer : Frequency Distribution Table Blood group O is most common as it has highest frequency, i.e., 12. Blood group AB is rarest as it has lowest frequency, i.e., 3.

A recent survey found that the ages of workers in a factory are distributed as follows: -Maths 9th

Description : A recent survey found that the ages of workers in a factory are distributed as follows: -Maths 9th

Last Answer : Total number of workers = 38 + 27 + 86 + 46 + 3 = 200 (i) P (person is 40 years or more) = P (Person having age 40 to 49 years) + P (person having age 50 to 59 years) + P (person having age 60 and ... 40 to 19 years) + P (person having age 50 to 59 years) = 86/200 + 46/200 = 132/200 = 0.66

An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Description : An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Last Answer : (i) From figure , it is clear that angle between the lines joining stalls C,D and stalls C, B is ∠BCD . Given, ∠DBC = 60° and ∠BAC = 40° CD is a chord of circle. Here, ∠CBD and ... for society. (iii) Other social issues for which such compaigns are required, are old age home, orphanages, etc.

An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Description : An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Last Answer : (i) From figure , it is clear that angle between the lines joining stalls C,D and stalls C, B is ∠BCD . Given, ∠DBC = 60° and ∠BAC = 40° CD is a chord of circle. Here, ∠CBD and ... for society. (iii) Other social issues for which such compaigns are required, are old age home, orphanages, etc.

A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. -Maths 9th

Description : A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. -Maths 9th

Last Answer : Let P (A) = Probability of selecting a fair complexioned person. ThenP(A) = \(rac{20}{80}\) = \(rac{1}{4}\)Let P(B) = Probability of selecting a rich person. Then P(B) = \(rac{10}{80}\) = \(rac{1}{8}\)Let P (C) = ... ) = \(rac{1}{4}\)x \(rac{1}{8}\)x \(rac{5}{16}\) = \(rac{5}{512}\) = 0.009.

A bag contains 2n + 1 coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. -Maths 9th

Description : A bag contains 2n + 1 coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. -Maths 9th

Last Answer : (a) 10As (n + 1) coins are fair P (Tossing a tail) = \(rac{rac{n+1}{2}}{2n+1}\) = \(rac{n+1}{2(2n+1)}\)∴ P (Tossing a head) = 1 - \(rac{n+1}{2(2n+1)}\) = \(rac{4n+2-n-1}{2(2n+1)}\) = \(rac{3n+1}{4n+2}\)Given, \(rac{3n+1}{4n+2}\) = \(rac{31}{42}\)⇒ 126n + 42 = 124n + 62 ⇒ 2n = 20 ⇒ n = 10.