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

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
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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 (B ∩ C) = P ({TH}) = \(rac{1}{4}\), (A ∩ B ∩ C) = ϕ∴ P (A ∩ B ∩ C) = P (ϕ) = 0 ≠ P (A). P (B). 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
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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

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

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

Last 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 ... (C), i.e., if the events are pairwise independent and (ii) P (A ∩ B ∩ C) = P (A) . P (B) . P (C)

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

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

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.

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

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

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

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 : 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

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) = \(rac{n(A)}{n(S)}\) = \(rac{3}{4}.\)

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.

Three coins are tossed 100 times, and three heads one head occurred 14 times and head did not occur 23 times. Find the probability of getting more tha

Description : Three coins are tossed 100 times, and three heads one head occurred 14 times and head did not occur 23 times. Find the probability of getting more tha

Last Answer : Three coins are tossed 100 times, and three heads one head occurred 14 times and head did not ... Find the probability of getting more than one head.

A beggar found a leather purse that someone had dropped in the marketplace. Opening it, he discovered that it contained 100 pieces of gold. Then he heard a merchant shout, 'A reward! A reward to the one who finds my leather purse!' Being an honest man, the beggar came forward and handed the purse to the merchant saying, 'Here is your purse. May I have the reward now?' 'Reward?' scoffed the merchant, greedily counting his gold. 'Why the purse I dropped had 200 pieces of gold in it. You've already stolen more than the reward! Go away or I'll tell the police.' 'I'm an honest man,' said the beggar defiantly. 'Let us take this matter to the court.' In court, the judge patiently listened to both sides of the story of a leather bag lost and a leather bag found. He counted the coins; 100 gold coin

Description : A beggar found a leather purse that someone had dropped in the marketplace. Opening it, he discovered that it contained 100 pieces of gold. Then he heard a merchant shout, 'A reward! A reward to the one ... of a leather bag lost and a leather bag found. He counted the coins; 100 gold coin -Riddles

Last Answer : The judge said 'Merchant, you said that the purse you lost contained 200 peices of gold. The purse this beggar found only contained 100 peices of gold. Therefore it cannot be the same purse'. With that the judge gave the purse to the beggar.

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.

What is the probability that the first two flips will both be heads and the third flip will be tails if you flip three fair coins?

Description : What is the probability that the first two flips will both be heads and the third flip will be tails if you flip three fair coins?

Last Answer : Need answer

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 ______.

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

Two coins are tossed. Find the number of outcomes of getting one head.

Description : Two coins are tossed. Find the number of outcomes of getting one head.

Last Answer : Two coins are tossed. Find the number of outcomes of getting one head.

When 2 coins are tossed simultaneously, write all possible outcomes.

Description : When 2 coins are tossed simultaneously, write all possible outcomes.

Last Answer : When 2 coins are tossed simultaneously, write all possible outcomes.

If we tossed simultaneously two coins. Find the probability of exactly one tail.

Description : If we tossed simultaneously two coins. Find the probability of exactly one tail.

Last Answer : If we toss two coins simultaneously,there are four possible outcomes HEAD-HEAD  TAIL-TAIL HEAD-TAIL  TAIL-HEAD  so probability of getting exactly one tail=2/4=1/2

Let us consider emission of `alpha`particle from uranium nucleus: `._(92)^(235)U - ._(2)He^(4) to ._(90)Th^(231)` `{:(e = 92, e = 0, e = 90),(p = 92,

Description : Let us consider emission of `alpha`particle from uranium nucleus: `._(92)^(235)U - ._(2)He^(4) to ._(90)Th^(231)` `{:(e = 92, e = 0, e = 90),(p = 92,

Last Answer : Let us consider emission of `alpha`particle from uranium nucleus: `._(92)^(235)U - ._( ... evolve energy C. annihilation D. absorption in the nucleus

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.

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\(rac{1}{2}\) = \(rac{1}{4}\).

Hari has some two rupee and five rupee coins .The total amount with him is rs. 43. Express the given information as a linear equation in two variables. -Maths 9th

Description : Hari has some two rupee and five rupee coins .The total amount with him is rs. 43. Express the given information as a linear equation in two variables. -Maths 9th

Last Answer : answer:

If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Description : If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Last Answer : (a) 0Let h be the first term and k be the common ratio of a GP, then a = hkp - 1, b = hkq - 1, c = hkr - 1∴ (q - r) log a + (r - p) log b + (p - q) log c = log [hkp -1]q - r + log [hkq -1]r - p + log[hkr -1]p - ... r + r - p + p - q) (kp - 1)q - r (kq -1)r - p (kr -1)p - q = log(ho ko) = log 1 = 0.

A cylinder is filled to 4/5 th of its volume. It is, then tilted so that the level of water coincides with one edge of its bottom and top edge -Maths 9th

Description : A cylinder is filled to 4/5 th of its volume. It is, then tilted so that the level of water coincides with one edge of its bottom and top edge -Maths 9th

Last Answer : answer:

Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Description : Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Last Answer : Let P(1, -1), Q \(\big(rac{-1}{2},rac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.Then, PQ = \(\sqrt{\big(rac{-1}{2}-1\big)^2+\big(rac{1}{2}+1\big)^2}\) = \(\sqrt{rac{9}{4}+rac{9}{4}} ... {3\sqrt2}{2}\)PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3∵ PQ = QR, the triangle PQR is isosceles.

A man flipped a coin 9 times and every single time it landed on heads. If the man flipped the coin again, what is the chance that it lands on heads? -Riddles

Description : A man flipped a coin 9 times and every single time it landed on heads. If the man flipped the coin again, what is the chance that it lands on heads? -Riddles

Last Answer : 50%. It's always 50%.

if you flipped a coin 60 times how many times would you expect the coin to land on heads?

Description : if you flipped a coin 60 times how many times would you expect the coin to land on heads?

Last Answer : 30 maybe but i say 35 or 31

If You toss a coin five times and it lands heads up each time. What is the probability that it will land heads up on the six toss Explain?

Description : If You toss a coin five times and it lands heads up each time. What is the probability that it will land heads up on the six toss Explain?

Last Answer : Need answer

If you flip a coin and roll a 6-sided die what is the probability that you will flip a heads and roll at least a 3?

Description : If you flip a coin and roll a 6-sided die what is the probability that you will flip a heads and roll at least a 3?

Last Answer : one in three1 2 = under 33 4 5 6 = at least 3so 4/6 probabilitycoin is 1 in 24/6 x 1/2 = 4/12 = 1/3

If you flip a coin and roll a 6-sided die what is the probability that you will flip a heads and roll at least a 3?

Description : If you flip a coin and roll a 6-sided die what is the probability that you will flip a heads and roll at least a 3?

Last Answer : one in three1 2 = under 33 4 5 6 = at least 3so 4/6 probabilitycoin is 1 in 24/6 x 1/2 = 4/12 = 1/3

In a probability experiment Eric flipped a coin 36 times. The coin landed on heads 24 times. What is the ratio of heads to tails in this experiment?

Description : In a probability experiment Eric flipped a coin 36 times. The coin landed on heads 24 times. What is the ratio of heads to tails in this experiment?

Last Answer : 2 to 1

Assertion (A): The coin when flipped next time will come up tails. Reason (R): Because the coin was flipped five times in a row, and each time it came up heads. Choose the correct answer from below: (A) Both (A) and (R) are true, and (R) is the correct explanation of (A). (B) Both (A) and (R) are false, and (R) is the correct explanation of (A). (C) (A) is doubtful, (R) is true, and (R) is not the correct explanation of (A). (D) (A) is doubtful, (R) is false, and (R) is the correct explanation of (A).

Description : Assertion (A): The coin when flipped next time will come up tails. Reason (R): Because the coin was flipped five times in a row, and each time it came up heads. Choose the correct answer from below: (A) Both (A ... A). (D) (A) is doubtful, (R) is false, and (R) is the correct explanation of (A).

Last Answer : (C) (A) is doubtful, (R) is true, and (R) is not the correct explanation of (A).

1 , 2 , 5 Money Coins These Why Government Coin ?

Description : 1 , 2 , 5 Money Coins These Why Government Coin ?

Last Answer : We Everyone I know 1 , 2 , 5 Money Coins Government Coin Because In these Money Secretary Signature From And The rest In the notes Stays Bangladesh Of the bank Of the governor . Basically Of the ... The reins Pull Caught Goes . That's why Extensive Used All this Coins Government Coin To say Is.

What is the number of each type of coin if A lady has rupees 1 and rupees 2 coins in her purse if in all she has 50 coins totalling rupees 70?

Description : What is the number of each type of coin if A lady has rupees 1 and rupees 2 coins in her purse if in all she has 50 coins totalling rupees 70?

Last Answer : Need answer