Question

An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is :

• A. $\frac{255}{256}$
• B. $\frac{127}{128}$
• C. $\frac{63}{64}$
• D. $\frac{1}{2}$

Answer 1

The Correct Option is B

To solve this problem, we need to find the probability of getting at least one head and at least one tail when an unbiased coin is tossed eight times.

We can approach this by using the concept of complementary probability. The complementary events to getting at least one head and at least one tail are:

1. All coins show heads.
2. All coins show tails.

First, we calculate the probability of the coin showing the same face (either all heads or all tails) in all eight tosses:

Probability of all heads: The probability of a head in one toss is \(\frac{1}{2}\). Thus, for eight tosses:

\(P(\text{All Heads}) = \left(\frac{1}{2}\right)^8 = \frac{1}{256}\)

Probability of all tails: Similarly, the probability for tails is also:

\(P(\text{All Tails}) = \left(\frac{1}{2}\right)^8 = \frac{1}{256}\)

Since these two events (all heads or all tails) are mutually exclusive, the probability of either all heads or all tails occurring is the sum of their probabilities:

\(P(\text{All Heads or All Tails}) = \frac{1}{256} + \frac{1}{256} = \frac{2}{256} = \frac{1}{128}\)

Complementary Probability: Now, we find the probability of the complement, which is obtaining at least one head and at least one tail:

\(P(\text{At least one head and one tail}) = 1 - P(\text{All Heads or All Tails})\)

\(= 1 - \frac{1}{128} = \frac{128}{128} - \frac{1}{128} = \frac{127}{128}\)

Therefore, the probability of obtaining at least one head and at least one tail is \(\frac{127}{128}\).

The correct answer is: \(\frac{127}{128}\).