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 your answer as a common fraction.

I'm confused on this problem. I don't even know what the problem is asking. :D

The possible outcomes were

HHt
Htt
ttt

So, the probability of at least one head is 2/3

6/7 because 7/8 and 6/7

No problem! Let's break down the problem step by step to better understand it.

First, let's consider the fact that three quarters are being tossed. This means that we have three independent events: the first coin toss, the second coin toss, and the third coin toss.

Now, the problem states that at least one tail appears. This means that there are two possibilities to consider:
1) One tail and two heads
2) Three tails

However, we are only interested in the probability that at least one head appears. So, we can disregard the second possibility of three tails and focus on the first possibility.

To calculate the probability of one tail and two heads, we need to consider the probability of each independent event.

The probability of getting a head on any single coin toss is 1/2, and the probability of getting a tail is also 1/2. Since these are independent events, the probabilities multiply together.

Now, let's calculate the probability of getting one tail and two heads:

P(one tail and two heads) = P(tail) * P(head) * P(head)

= (1/2) * (1/2) * (1/2)

= 1/8

Therefore, the probability of getting at least one head is equal to 1 - P(one tail and two heads).

P(at least one head) = 1 - P(one tail and two heads)

= 1 - 1/8

= 7/8

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

I hope this explanation helps! Feel free to ask if you have any further questions.