Question

A fair coin is tossed 4 times. The probability that at least one head occurs is

• A. \(\frac{1}{4}\)
• B. \(\frac{7}{8}\)
• C. \(\frac{6}{8}\)
• D. \(\frac{15}{16}\)

Hint:

When dealing with "at least one" type of probability problems, it's often simpler to calculate the probability of the opposite (none occur) and subtract it from 1. This avoids enumerating all the cases where at least one head occurs, which can be more numerous.

Answer 1

The Correct Option is D

Step 1: Understand the sample space of tossing a fair coin 4 times.
When a fair coin is tossed once, there are two possible outcomes: Head (H) or Tail (T). When it is tossed 4 times, the total number of possible outcomes is \(2 \times 2 \times 2 \times 2 = 2^4 = 16\). These outcomes are equally likely since the coin is fair. The sample space (S) consists of all possible sequences of 4 tosses: S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
The total number of outcomes in the sample space is \(|S| = 16\). Step 2: Define the event of interest.
We are interested in the probability that at least one head occurs. Let E be the event that at least one head occurs. Step 3: Calculate the probability of the complementary event.
It is often easier to calculate the probability of the complementary event, which is the event that no heads occur. If no heads occur in 4 tosses, it means all 4 tosses resulted in tails. There is only one such outcome: TTTT. Let \(E^c\) be the complementary event that no heads occur (i.e., all tails).
The number of outcomes in \(E^c\) is \(|E^c| = 1\). The probability of the complementary event \(P(E^c)\) is the number of favorable outcomes divided by the total number of possible outcomes: $$P(E^c) = \frac{|E^c|}{|S|} = \frac{1}{16}$$ Step 4: Use the relationship between the probability of an event and its complement.
The probability of an event \(E\) is given by:
$$P(E) = 1 - P(E^c)$$
Substitute the value of \(P(E^c)\) we calculated:
$$P(E) = 1 - \frac{1}{16} = \frac{16}{16} - \frac{1}{16} = \frac{15}{16}$$ Therefore, the probability that at least one head occurs is \(\frac{15}{16}\). Step 5: Match the result with the given options.
The calculated probability \(\frac{15}{16}\) matches option (4).