In the experiment of tossing a fair coin 3 times, the sample space is made of the
following 8 equiprobable events: S = {HHH, THH, HTH, HHT, TTH, THT, HTT, TTT}.
The events that at least have one head are 7. So the probability of getting at least
one head on three tosses is: P(at least one H) = 7/8 = 0.875 = 87.5%
0.53 = 0.125.
If you toss the die often enough then the probability of getting the sequence 2-2-1 is 1: a certainty. The probability of getting the result in the first three tosses is 1/216.
The probability of getting heads on three tosses of a coin is 0.125. Each head has a probability of 0.5. Since the events are sequentially unrelated, simply raise 0.5 to the power of the number of tosses (3) and get 0.125, or 1 in 8.
In three tosses, the probability is 3/8.
It is 1/8.
The probability is 0.0322
33%
0.53 = 0.125.
It is 3/8.
2 out of 3 i think
In a large enough number of tosses, it is a certainty (probability = 1). In only the first three tosses, it is (0.5)3 = 0.125
Pr(At least one head in three tosses) = 1 - Pr(No heads in three tosses) = 1 - Pr(Three tails in three tosses) = 1 - (1/2)*(1/2)*(1/2) = 1 - 1/8 = 7/8 or 0.875 or 87.5%
If you toss the die often enough then the probability of getting the sequence 2-2-1 is 1: a certainty. The probability of getting the result in the first three tosses is 1/216.
It is 93/256 = 0.363 approx.
Pr(At least one head in 3 tosses) = 1 - Pr(No heads in 3 tosses) = 1 - Pr(3 tails in three tosses) = 1 - [Pr(T)*Pr(T)*Pr(T)] since the three tosses are independent. = 1 - 1/2 * 1/2 *1/2 = 1 - 1/8 = 7/8
>>> 1:7 (or, if you like probability, 87.5%)I disagree. There are four possible combinations of three tosses (where order does not matter):HHHHHTHTTTTTThree of these combinations will show at least one head - only by throwing three tails will you not throw at least one head.Thus, the probability of throwing at least one head in three flips is 75%.
In three tosses, the probability is 3/8.