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

1 Answer

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

Related questions

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

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

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

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

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

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

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

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

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

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

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

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

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.

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 : 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

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

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

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

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.

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

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?

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.

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 ______.

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

Description : Two dice are rolled simultaneously. The probability of getting a multiple of 2 on one dice and a multiple of 3 on the other is -Maths 9th

Last Answer : (c) \(rac{11}{36}\)Total number of outcomes when two identical dice are rolled, n(S) = 6 6 = 36 Let A : Event of rolling a multiple of 2 on one die and a multiple of 3 on the other die ⇒ A = {(2, 3), (2, 6), (4, 3), (4, ... , 4), (3, 6)} ⇒ n(A) = 11 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{11}{36}\).

Description : Two dice are rolling simultaneously .What is the probability that the sum of the number on the two faces is divided by 5 Or 7. A) 13/36 B) 14/36 C) 11/36 D) 9/36 

Last Answer :  Answer: C) Clearly, n(S) = 6 x 6 = 36 Let E be the event that the sum of the numbers on the two faces is divided by 5or 7. Then,E = {(1,4),(1,6),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(4,6),(5,2),(5,5),(6,1),(6,4)} n(E) = 11. Hence, P(E) = n(E)/n(S) = 11/36

Description : For an unbiased coin, the probability of getting tail is 1/2.

Last Answer : For an unbiased coin, the probability of getting tail is 1/2.

Description : How do you make 1 dollar using pennies nickels dimes and quarters using exactly 50 coins?

Last Answer : There are 2 solutions (if you include the non-use of quarters): 1 Quarter, 2 Dimes, 2 Nickels, 45 Pennies No Quarters, 2 Dimes, 8 Nickels, 40 Pennies

Description : The Matrix has Complex eigenvalues and vectors

Last Answer : The Matrix has Complex eigenvalues and vectors

Description : For the given orthogonal Matrix, the inverse is,

Last Answer : For the given orthogonal Matrix, the inverse is,

Description : determinant of 2 by 2 matrix

Last Answer : determinant of 2 by 2 matrix

Description : Linear algebra problem: Rank of a matrix

Last Answer : Linear algebra problem: Rank of a matrix https://youtu.be/b7tPLA3_y6Y

Description : Linear algebra problem: The determinant of the matrix

Last Answer : Answer : Linear algebra problem: The determinant of the matrix

Description : The lowest eigenvalue of the 2x2 matrix 

Last Answer : Linear Algebra Problem The lowest eigenvalue of the 2x2 matrix https://youtu.be/QatcRTzF7EA

Description : Linear Algebra Problem the solution to the system of equations

Last Answer : Linear Algebra Problem the solution to the system of equations https://youtu.be/y_FlupirxZM

Description : Problem on eigen values property

Last Answer : Problem on eigen values property https://youtu.be/DoSTY-bBubQ

Description : The determinant of a 2 x 2 matrix is 50. If one eigen value of the matrix is 10, the other eigen value is ________.

Last Answer : https://youtu.be/1WrgXbDX1K4

Description : Linear Algebra Problem 2

Last Answer : https://youtu.be/zAPDYO0q-9w Linear Algebra Problem 2

Description : Linear Algebra Problem The product of eigen values of the matrix

Last Answer : Linear Algebra Problem The product of eigen values of the matrixhttps://youtu.be/JGcynUZFQ78

Description : Can you answer this question? X + X + X = 30 X + Y + Y = 20 Y + Z + Z = 17 X + Y + Z = ?

Last Answer : X+Y+Z =21 10+05+6=21

Description : 5(4+1) ÷ 5(5) = ?

Last Answer : 5(4+1) 5(5) = 20/25 You can simplify the expression by first performing the operations inside the parentheses. So, 5(4+1) becomes 5*5 which is 25. Then 25 5( ... of operations is important, you have to perform the operations inside the parentheses first, then the multiplication and division.

Description : Find the area of rhombus whose diagonals are 6 cm and 9 cm.

Last Answer : Area of rhombus = 1/2×d1×d2 = 1/2×6×9 = 27 ans

Description : Find the value of (sin 15o ) using compound angles.

Last Answer : Solution:

Description : Find the area of the triangle whose vertices are 4,3 1,4 and 2,3 . 

Last Answer : solution:

Description : Evaluate log3 81.

Last Answer : Solution: