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What is the probability of obtaining exactly 4 heads in 6 flips of a fair coin?
Wiki User
∙ 13y ago
The probability of a head in one flip is 1/2.
The probability of HHHHTT is (1/2)6 = 1/64
The possible correct flips are HHHHTT, HHHTHT, HHTHHT, HTHHHT, THHHHT, HHHTTH, HHTHTH, HTHHTH, THHHTH, HHTTHH, HTHTHH, THHTHH, HTTHHH, THTHHH, TTHHHH, each with a probability of 1/64.
Total probability is 15/64.
Wiki User
∙ 13y ago
I have dyscalculia
Lesson: Normal Distribution
Problem Solving
A. Given a standard normal distribution, determine the are of the following regions?
a. to the left of z=1.25
b. to the right of z=0.06
c. between z=-1.5 and z=1.7
Anonymous
Ashley knows that h3 =343. Which of the following expression will give Ashley the correct value of h?C.343 3
Anonymous
Anonymous
1/2 probability
Anonymous
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What is the probability of obtaining exactly seven heads in eight flips of a coin given that at least one is a head?
The probability of obtaining 7 heads in eight flips of a coin is:P(7H) = 8(1/2)8 = 0.03125 = 3.1%
What is the probability of obtaining exactly six heads in seven flips of a coin given that at least one is a head?
The requirement that one coin is a head is superfluous and does not matter. The simplified question is "what is the probability of obtaining exactly six heads in seven flips of a coin?"... There are 128 permutations (27) of seven coins, or seven flips of one coin. Of these, there are seven permutations where there are exactly six heads, i.e. where there is only one tail. The probability, then, of tossing six heads in seven coin tosses is 7 in 128, or 0.0546875.
What is the probability of obtaining exactly four heads in five flips of a coin given that at least three are heads?
We can simplify the question by putting it this way: what is the probability that exactly one out of two coin flips is a head? Our options are HH, HT, TH, TT. Two of these four have exactly one head. So 2/4=.5 is the answer.
Maria flips a dime four times What is the probability that she will get at least two heads?
It is 0.6875
What is the probabilty that a coin will land on heads if flipped 8 times?
This is a probability question. Probabilities are calculated with this simple equation: Chances of Success / [Chances of Success + Chances of Failure (or Total Chances)] If I flip a coin, there is one chance that it will land on heads and one chance it will land on tails. If success = landing on heads, then: Chances of Success = 1 Chances of Failure = 1 Total Chances = 2 Thus the probability that a coin will land on heads on one flip is 1/2 = .5 = 50 percent. (Note that probability can never be higher than 100 percent. If you get greater than 100 you did the problem incorrectly) Your question is unclear whether you mean the probability that a coin will land on head on any of 8 flips or all of 8 flips. To calculate either you could write out all the possible outcomes of the flips (for example: heads-heads-tails-tails-heads-tails-heads-heads) but that would take forvever. Luckily, because the outcome of one coin flip does not affect the next flip you can calculate the total probability my multiplying the probabilities of each individual outcome. For example: Probability That All 8 Flips Are Heads = Prob. Flip 1 is Heads * Prob. Flip 2 is Heads * Prob. Flip 3 is Heads...and so on Since we know that the probability of getting heads on any one flips is .5: Probability That All 8 Flips Are Heads = .5 * .5 * .5 * .5 * .5 * .5 * .5 * .5 (or .58) Probability That All 8 Flips Are Heads = .00391 or .391 percent. The probability that you will flip a heads on any of flips is similar, but instead of thinking about what is the possiblity of success, it is easier to approach it in another way. The is only one case where you will not a heads on any coin toss. That is if every outcome was tails. The probability of that occurring is the same as the probability of getting a heads on every toss because the probability of getting a heads or tails on any one toss is 50 percent. (If this does not make sense redo the problem above with tails instead of heads and see if your answer changes.) However this is the probability of FAILURE not success. This is where another probability formula comes into play: Probability of Success + Probability of Failure = 1 We know the probability of failure in this case is .00391 so: Probability of Success + .00391 = 1 Probability of Success = .9961 or 99.61 percent. Therefore, the probability of flipping a heads at least once during 8 coin flips is 99.61 percent. The probability of flipping a heads every time during 8 coin flips is .391 percent.
What is the probability of obtaining exactly 5 heads in 6 flips of a fair coin?
It is approx 0.0938
What is the probability of obtaining exactly seven heads in eight flips of a coin given that at least one is a head?
The probability of obtaining 7 heads in eight flips of a coin is:P(7H) = 8(1/2)8 = 0.03125 = 3.1%
What is the probability of obtaining exactly six heads in seven flips of a coin given that at least one is a head?
The requirement that one coin is a head is superfluous and does not matter. The simplified question is "what is the probability of obtaining exactly six heads in seven flips of a coin?"... There are 128 permutations (27) of seven coins, or seven flips of one coin. Of these, there are seven permutations where there are exactly six heads, i.e. where there is only one tail. The probability, then, of tossing six heads in seven coin tosses is 7 in 128, or 0.0546875.
What is the probability of obtaining exactly six heads in seven flips of a coin?
7*(1/2)7 = 7/128 = 5.47% approx.
What is the probability of obtaining exactly four heads in five flips of a coin given that at least three are heads?
We can simplify the question by putting it this way: what is the probability that exactly one out of two coin flips is a head? Our options are HH, HT, TH, TT. Two of these four have exactly one head. So 2/4=.5 is the answer.
What is the probability of obtaining exactly two heads in three flips of a coin given that at least one is a head?
The probability of obtaining exactly two heads in three flips of a coin is 0.5x0.5x0.5 (for the probabilities) x3 (for the number of ways it could happen). This is 0.375. However, we are told that at least one is a head, so the probability that we got 3 tails was impossible. This probability is 0.53 or 0.125. To deduct this we need to divide the probability we have by 1-0.125 0.375/(1-0.125) = approximately 0.4286
What is the probability of exactly three heads in four flips of a coin given at least two are heads?
If you know that two of the four are already heads, then all you need to find isthe probability of exactly one heads in the last two flips.Number of possible outcomes of one flip of one coin = 2Number of possible outcomes in two flips = 4Number of the four outcomes that include a single heads = 2.Probability of a single heads in the last two flips = 2/4 = 50%.
What is the probability of obtaining 3 heads in three flips of a fair coin?
1/2 * 1/2 * 1/2 = 1/8 = 12.5%
What is the probability of obtaining exactly three heads in four flips of a coin given that at least two are heads?
Pr(3H given >= 2H) = Pr(3H and >= 2H)/Pr(>=2H) = Pr(3H)/Pr(>=2H) = (1/4)/(11/16) = 4/11.
What is the probability of obtaining exactly 4 heads in a coin?
About a 1 in 16 chance of getting a coin to land on heads 4 times in a row.
What is the probability of obtaining exactly three heads in four flips of a coin given that at least one is a head?
50-50. each toss is independent of any previous toss. if all tosses are to be heads/tails then each toss you multiply by the number of chances. i,e. 2, starting with 1. three heads in a row is 1x2x2
What is the probability that in four coin flips you get at least 2 heads?
50%
