Three fair 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.
There are a total of eight possible outcomes for three coins: HHH, HHT, HTH, THH, TTH, THT! HTT, TTT. Out of these eight outcomes, seven of these show at least one tail (all except HHH). After counting, we find that six of these seven outcomes have at least one head, so the probability is—6/7.
6/7
To solve this problem, we can use conditional probability. Let's break it down step by step.
Step 1: Determine the total number of possible outcomes. Since there are 3 fair quarters being tossed, each having 2 possible outcomes (head or tail), the total number of possible outcomes is 2^3 = 8.
Step 2: Determine the number of outcomes where at least one tail appears. To find this number, we can subtract the number of outcomes with all heads from the total number of possible outcomes. The only outcome where all three quarters show heads is when all three tosses result in heads. Therefore, there is only one outcome without a tail. So, the number of outcomes with at least one tail is 8 - 1 = 7.
Step 3: Determine the number of outcomes where at least one head appears. Since we are looking for the complementary event to the previous condition (i.e., at least one head instead of at least one tail), we can subtract the number of outcomes without a head from the total number of possible outcomes. Again, the only outcome without a head is when all three tosses result in tails. So, the number of outcomes with at least one head is 8 - 1 = 7.
Step 4: Calculate the probability. The probability that at least one head appears, given that at least one tail appears, is the number of outcomes with both conditions satisfied (at least one head and at least one tail) divided by the number of outcomes with at least one tail. Therefore, the probability is 7/7 = 1.
So, the probability that at least one head appears, given that at least one tail appears, is 1, or 1/1 as a common fraction.
For the other two, the probablilty of two tails is 1/2*1/2=1/4
Therefore, pr(at least one head)=1-1/4=3/4