Four fair coins are tossed simultaneously. The probability that at least one tail turns up is:
Show Hint
- \( \frac{1}{16} \)
- \( \frac{15}{16} \)
- \( \frac{7}{8} \)
- \( \frac{1}{2} \)
The Correct Option is B
Solution and Explanation
We are asked to find the probability that at least one tail turns up when four fair coins are tossed.
First, we need to calculate the total number of possible outcomes. Since each coin has two possible outcomes (Heads or Tails), and there are four coins, the total number of possible outcomes is:
\[ 2^4 = 16 \]
Next, let’s consider the complement of the event we are interested in, which is the event that no tails show up (i.e., all coins show heads).
The number of outcomes where no tails show up is just one — the outcome where all four coins land heads:
\[ \text{Number of outcomes with no tails} = 1 \]
Now, we can calculate the number of outcomes where at least one tail appears.
This is the complement of the event where no tails show up:
\[ \text{Number of outcomes with at least one tail} = 16 - 1 = 15 \]
Therefore, the probability of having at least one tail is the ratio of favorable outcomes (15) to total outcomes (16):
\[ P(\text{at least one tail}) = \frac{15}{16} \]
Thus, the probability that at least one tail turns up is \( \frac{15}{16} \).
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