Want this question answered?

Q: Assuming the sequences are all equally likely what is the probability that you will get exactly two heads when you toss a coin three times?

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

The answer is 1/2 , assuming the coin is fair.

The probability of exactly 3 girls in a family of 10 children, assuming equal chance of a boy or girl, is 0.1172. This is a binomial distribution.

P(3a) = 5c3 ∙(0.61)3∙(0.39)2 = 0.3452... ~ 34.5%

The probability is 2 - 6

The probability is 0.375

Related questions

50-50

A computer is programmed to generate a sequence of three digits, where each digit is either 0 or 1, and each of these is equally likely to occur. Construct a sample space that shows all possible three-digit sequences of 0s and 1s and then find the probability that a sequence will contain exactly one 0.

A computer is programmed to generate a sequence of three digits, where each digit is either 0 or 1, and each of these is equally likely to occur. Construct a sample space that shows all possible three-digit sequences of 0s and 1s and then find the probability that a sequence will contain exactly one 0.

The answer is 1/2 , assuming the coin is fair.

The probability of exactly 3 girls in a family of 10 children, assuming equal chance of a boy or girl, is 0.1172. This is a binomial distribution.

P(3a) = 5c3 ∙(0.61)3∙(0.39)2 = 0.3452... ~ 34.5%

Assuming we want two tails exactly, the possible options to get them are: TTH, THT and HTT. They are three choices out of the eight available, which is a probability of 3/8, 0.375 or 37.5%.

The probability is 0.375

The probability is 2 - 6

If you mean 'at least' 2 heads, the probability is 50%. If you mean exactly 2, the probability is 3/8, or 37.5%. There are 3 independent coin tosses, each of which is equally likely to come up heads or tails. That's a total of 2 * 2 * 2 or 8 possible outcomes (HHH, HHT, HTH, etc.). Of these, 4 include 2 or 3 heads, which is half of 8. Only 3 include exactly 2 heads, so the probability of that is 3/8.

The probability this student will fail is the same as the probability that some other student will flip a fair coin 20 times and get less than 8 heads, i.e., more than 12 tails. There are 2^20 possible different-looking sequences of 20 coinflips, which we assume all have equal probability. Of those sequences, 1 has no heads at all, 20 have exactly 1 head, 190 have exactly 2 heads, ... and 77520 have exactly 7 heads. So we sum up all those possible ways to fail and we get ... ... ... I'm assuming that the student answers randomly, flipping a fair (50:50) coin on each question to choose "true" or "false". In that special case, it doesn't matter how many of those twenty questions are true or how many are false. (If the student answers randomly by flipping an unfair coin, say a 25:75 coin, then it does matter how many of those questions are true -- I'll let you figure that one out).

The probability of getting exactly eight heads when tossing 10 coins once can be found using the binomial probability formula. Assuming a fair coin, the probability of getting a heads is 1/2. Plugging in the numbers, the probability of getting exactly eight heads is (10 choose 8) * (1/2)^8 * (1/2)^2 = 45/1024, which is approximately 0.04395.