Two coins are tossed. What is the probability of getting at most one head?

Probability of Getting at Most One Head in Two Coin Tosses

When two coins are tossed, we need to determine the probability of getting at most one head. Let’s break this problem down step by step.

Sample Space:

The sample space S of tossing two coins consists of all possible outcomes. Each coin has two faces, heads (H) and tails (T). Therefore, the sample space is:

S = \{ (H,H), (H,T), (T,H), (T,T) \}

There are a total of 4 possible outcomes when two coins are tossed.

Favorable Outcomes

We are looking for the probability of getting at most one head. This includes the following scenarios:

• Getting zero heads

• Getting one head

The favorable outcomes that meet this criterion from our sample space S are:

\{ (T,T), (H,T), (T,H) \}

Thus, there are 3 favorable outcomes.

Calculating the Probability

The probability P(A) of an event A is given by the ratio of the number of favorable outcomes to the total number of possible outcomes:

P(A) = \frac{ \text{Number of favorable outcomes} }{ \text{Total number of possible outcomes} }

For our problem, this translates to:

$P(\text{At most one head}) = $\frac{ 3 }{ 4 }$

Answer:

Therefore, the probability of getting at most one head when two coins are tossed is: 3/4.

Verification:

Let’s verify by considering each probability individually:

• Probability of getting zero heads (TT): \frac{1}{4}

• Probability of getting one head (HT or TH): $\frac{2}{4} = \frac{1}{2}$

Adding these probabilities:

$\frac{1}{4} + \frac{2}{4} = \frac{3}{4}$

This confirms our calculation is correct.

The correct answer is indeed 3/4.