The answer depends on what the experiment is!
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
The probability is 5/16.
The answer depends on what the experiment is!
It is approx 0.0938
The answer depends on what the experiment is!
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
The probability is 5/16.
1/2 * 1/2 * 1/2 = 1/8 = 12.5%
The probability of obtaining 7 heads in eight flips of a coin is:P(7H) = 8(1/2)8 = 0.03125 = 3.1%
The answer depends on what the experiment is!
It is approx 0.0938
Mr. Palmer moved to Canada in 1974 where he continued to teach and write. He died there on June 16, 2013. Related Documents.
About a 1 in 16 chance of getting a coin to land on heads 4 times in a row.
The opposite of getting at most two heads is getting three heads. The probability of getting three heads is (1/2)^2, which is 1/8. The probability of getting at most two heads is then 1 - 1/8 which is 7/8.
The probability is 0.09766%.Each toss has a ½ chance to be heads. To combine probabilities use multiply them. So the probability to get two heads out of two tosses is ½ * ½, and three heads out of three tosses is ½ * ½ * ½. So the exact answer is 0.5^10
Assume the given event depicts flipping a fair coin and rolling a fair die. The probability of obtaining a tail is ½, and the probability of obtaining a 3 in a die is 1/6. Then, the probability of encountering these events is (½)(1/6) = 1/12.