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

1 Answer

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) = \(rac{n(A)}{n(S)}\) = \(rac{4}{8}\) = \(rac{1}{2}.\)

Related questions

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

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

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

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

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

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

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Last Answer : Need answer

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

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

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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

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

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

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

Description : Three coins are tossed simultaneously -Maths 9th

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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.

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

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

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Description : A die is rolled three times. The probability of getting a larger number than the previous number each time is: -Maths 9th

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

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

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Description : When you roll a 6 sided die with faces numbered 1 through 6 and toss a coin what is the probability of rolling a 5 and getting tails?

Last Answer : As the two events are independent multiply the respectiveprobabilities together:pr(5) = 1/6pr(tails) = 1/2*→ pr(5 and tails) = 1/6 1/2 = 1/12*Actually it is ... coin itbecomes a real possibility as (accidentally) demonstrated in the1997 Royal Institution Christmas Lecture "The Magical Maze"

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.

Description : In a throw of a die, find the probability of getting an even number. -Maths 9th

Last Answer : Total even number on a die = 3 P (getting an even numbers) = 3/6 = 1/2

Description : Two dice are thrown. Find the probability of getting an odd number on the first die and a multiple of 3 on the other. -Maths 9th

Last Answer : Let A : Getting an odd number on first die; B : Getting a multiple of 3 on second die, thenA = {1, 3, 5}, B = {3, 6} ∴ P(A) = \(rac{3}{6}=rac{1}{2}\), P(B) = \(rac{2}{6}=rac{1}{3}\) ... B are independent∴ Required probability = P (A) . P (B) = \(rac{1}{2}\) x \(rac{1}{3}\) = \(rac{1}{6}\)

Description : Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7. -Maths 9th

Last Answer : (c) \(rac{7}{12}\)Total number of ways in which 2 dice are rolled = 6 6 = 36 ⇒ n(S) = 36 Let A : Event of rolling an even number of 1st dice B : Event of rolling a total of 7 ⇒ A = {(2, 1), (2, 2) , (2, 6), (4 ... (rac{18}{36}\) + \(rac{6}{36}\) - \(rac{3}{36}\) = \(rac{21}{36}\) = \(rac{7}{12}\).

Description : In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Last Answer : Required probability = 1 – P(4) – P(5) =1- 1 / 6 - 1 / 6 = 4 / 6 = 2 / 3

Description : In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Last Answer : Required probability = 1 – P(4) – P(5) =1- 1 / 6 - 1 / 6 = 4 / 6 = 2 / 3

Description : A die is thrown.What is the probability of getting a multiple of 3 on the upper face ? -Maths 9th

Last Answer : Multiple of 3 on a die = 3, 6 ∴ P (a multiple of 3) = 2/6 = 1/3.

Description : In a single throw of two dice, what is the probability of getting a sum of 9? -Maths 9th

Last Answer : Outcomes with sum of 9 = { (3, 6), (4, 5), (5, 4), (6, 3) } P ( getting a sum of 9 is ) = 4/36 = 1/9

Description : Two unbiased dice are rolled. Find the probability of getting a multiple of 2 on one die and a multiple of 3 on the other die ? -Maths 9th

Last Answer : When two unbiased dice are rolled, the possible out comes are∴ n(S) = 36 Let A : getting a multiple of 2 on one die and a multiple of 3 on the other die. ⇒ A = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), ( ... (3, 6), (6, 2), (6, 4)} ⇒ n(A) = 11∴ P(A) = \(rac{n(A)}{n(S)} =rac{11}{36}.\)

Description : What is the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays in a non–leap year ? -Maths 9th

Last Answer : A non-leap year consists of 365 days. Therefore in a non-leap year there are 52 complete weeks and 1 day over which can be one of the seven days of the week. Possible outcomes n(S) = 7 = {Sunday, Monday, Tuesday, Wednesday, Thursday, ... ⇒ n(A) = 3∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{3}{7}.\)

Description : A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th

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Description : My body has a dozen heads or more, My tails don’t wag when you walk in the door. Count the ways you can hold me tight, Or use me for a special night! What Am I? -Riddles

Last Answer : A piggy bank.

Description : What has fifty heads and fifty tails? -Riddles

Last Answer : Fifty pennies.

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).

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.

Description : Three identical dice are rolled. The probability that same number will appear on each of them is -Maths 9th

Last Answer : (d) \(rac{1}{36}\)Total number of outcomes when three identical dice are rolled, n(S) = 6 6 6 = 216 Let A : Event of rolling same number or each dice ⇒ A = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), ... 6)} ⇒ n(A) = 6 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{6}{216}\) = \(rac{1}{36}\).