Question

A coin is tossed 5 times. The probability of getting atleast one head is:

• A. \(\frac{31}{32}\)
• B. \(\frac{15}{16}\)
• C. \(\frac{1}{32}\)
• D. \(\frac{63}{64}\)

Answer 1

The Correct Option is A

A coin is tossed 5 times. We need to find the probability of getting at least one head. To solve this problem, we will use the complementary probability approach, which is often simpler for "at least one" scenarios.

Let \( P(\text{at least one head}) \) be the probability of getting at least one head. The complement of this event is getting no heads at all, which means getting all tails.

Step 1: Calculate the probability of getting all tails when a coin is tossed 5 times.

The probability of getting a tail in one toss is \( \frac{1}{2} \).

So, the probability of getting all tails in 5 tosses is:
\( P(\text{all tails}) = \left( \frac{1}{2} \right)^5 = \frac{1}{32} \).

Step 2: Use the complement rule to find the probability of getting at least one head:

\( P(\text{at least one head}) = 1 - P(\text{all tails}) \)

Therefore, we have:
\( P(\text{at least one head}) = 1 - \frac{1}{32} = \frac{32}{32} - \frac{1}{32} = \frac{31}{32} \).

Hence, the probability of getting at least one head when a coin is tossed 5 times is \(\frac{31}{32}\).

Probability of all tails	\(\frac{1}{32}\)
Probability of at least one head	\(\frac{31}{32}\)