A coin is tossed twice.
What is the probability of tossing heads, and then tails, given that the coin has already shown heads in the first toss?
that would be P(tails) on the 2nd toss: 1/2
Hey Steve, thanks! I was wondering though why I would of gotten 1/4 when I did the math the first time?
P(heads) = 1/2
P(tails) = 1/2
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
P(heads) * P(tails) = 1/2 * 1/2 = 1/4
However, the toss of heads was already completed.
To find the probability of tossing heads and then tails, given that the coin has already shown heads in the first toss, we need to consider the sample space and the event of interest.
The sample space represents all possible outcomes when tossing a coin twice. In this case, the sample space is {HH, HT, TH, TT}, where H represents heads and T represents tails.
Since we know that the coin has already shown heads in the first toss, we can eliminate the possibility of getting tails in the first toss. Therefore, the reduced sample space becomes {HH, HT}.
The event of interest is tossing heads, and then tails. From the reduced sample space, we can see that there is only one outcome that satisfies this event, which is HT.
To find the probability, we need to divide the number of favorable outcomes (HT) by the total number of possible outcomes in the reduced sample space ({HH, HT}). Therefore, the probability of tossing heads, and then tails, given that the coin has already shown heads in the first toss is:
P(HT|H) = Number of favorable outcomes / Total number of outcomes
= 1 / 2
= 0.5
= 50%
So, the probability is 0.5 or 50%.