Three quarters are tossed and a tail appears on at least one of them. What is the probability that at least one head appears? Express as a common fraction.

Well basically, just count how many total combinations there are in the flips.

= HHH, HHT, HTH, THH, TTH, THT, HTT, TTT

Of those, we find 7 with a tail (we omit HHH from our list).
Now with that, we count the ones with at least one head.

So..our answer is 6/7

There are a total of eight possible outcomes for three coins: HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT. Of these eight outcomes, seven of them show at least one tail (all except HHH).

Counting, we find six of these seven outcomes have at least one head, so the probability is 6/7.

To find the probability that at least one head appears when tossing three quarters and getting a tail on at least one of them, we need to consider the sample space and the favorable outcomes.

The sample space consists of all the possible outcomes when tossing three quarters. Since each coin has two possible outcomes (heads or tails), the total number of outcomes is 2^3 = 8.

Now let's consider the favorable outcomes, which are the outcomes where at least one head appears. To determine this, we can subtract the outcomes where all three coins land as tails from the total number of outcomes.

There is only one outcome where all three coins land as tails (TTT). Therefore, the number of favorable outcomes is 8 - 1 = 7.

Therefore, the probability that at least one head appears is 7/8.

Hence, the probability is expressed as a common fraction, 7/8.

Ah, probabilities, the realm of uncertainty! Let's ponder this perplexing problem.

To find the probability of at least one head appearing when three quarters are tossed and a tail appears on at least one of them, we need to consider the different scenarios.

There are a total of 2^3 = 8 possible outcomes when tossing three coins. Those outcomes are:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Out of these eight outcomes, three have at least one tail: THT, TTH, and TTT.

But we need to find the probability of at least one head appearing. In other words, we need to subtract the probability of getting three tails (TTT) from the total probability of getting at least one tail (THT, TTH, TTT).

So, the probability of at least one head appearing is 3/8 - 1/8 = 2/8 = 1/4.

Therefore, the probability that at least one head appears is 1/4. Keep calm and carry on tossing those coins!

2/3, because there are 3 coin flips, but one is already tails, so there's only 2 more chances to get heads.