Coin Toss Probability: Heads More Than Tails

This problem asks us to determine the probability of a specific event occurring when a fair coin is tossed independently four times. The event we are interested in is "the number of times heads show up is more than the number of times tails show up".

Understanding Coin Toss Properties

A fair coin implies that the probability of landing on Heads (H) is equal to the probability of landing on Tails (T). Therefore, for a single toss:

• Probability of Heads, $P(H) = \frac{1}{2}$
• Probability of Tails, $P(T) = \frac{1}{2}$

The problem states that the coin is tossed independently four times. This means that the result of one toss does not influence the result of any other toss.

Total Possible Outcomes (Sample Space)

When a coin is tossed multiple times, the total number of possible outcomes can be calculated by raising the number of outcomes per toss (2 for H or T) to the power of the number of tosses. In this case, there are 4 tosses:

Total Number of Outcomes $= (\text{Number of outcomes per toss})^{\text{Number of tosses}} = 2^4 = 16$.

These 16 outcomes form our complete sample space, representing all possible sequences of heads and tails over four tosses (e.g., HHHH, HHHT, HHTH, HTHT, etc.).

Identifying Favorable Outcomes for the Event

We need to find the outcomes where the "number of times heads show up is more than the number of times tails show up". Let's denote the number of heads as $N_H$ and the number of tails as $N_T$. For four tosses, we know that $N_H + N_T = 4$.

We need to find pairs of ($N_H, N_T$) such that $N_H > N_T$. Let's examine all possible values for $N_H$ from 0 to 4:

• If $N_H = 0$, then $N_T = 4$. Is $0 > 4$? No.
• If $N_H = 1$, then $N_T = 3$. Is $1 > 3$? No.
• If $N_H = 2$, then $N_T = 2$. Is $2 > 2$? No.
• If $N_H = 3$, then $N_T = 1$. Is $3 > 1$? Yes. This is a favorable case.
• If $N_H = 4$, then $N_T = 0$. Is $4 > 0$? Yes. This is another favorable case.

So, the favorable outcomes occur when we get exactly 3 heads and 1 tail, or exactly 4 heads and 0 tails.

Counting the Number of Favorable Outcomes

To count how many different sequences correspond to these favorable cases, we use the binomial coefficient formula, which helps us find the number of ways to choose $k$ items from a set of $n$ items, denoted as $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Here, $n$ is the total number of tosses (4), and $k$ is the number of heads.

Case 1: Exactly 3 Heads ($N_H=3$)

The number of ways to get exactly 3 heads in 4 tosses is $\binom{4}{3}$:

$$\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times 1} = 4$$

The possible sequences for 3 heads and 1 tail are: HHHT, HHTH, HTHH, THHH.

Case 2: Exactly 4 Heads ($N_H=4$)

The number of ways to get exactly 4 heads in 4 tosses is $\binom{4}{4}$:

$$\binom{4}{4} = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} = \frac{4!}{4! \times 1} = 1$$

The only sequence for 4 heads and 0 tails is: HHHH.

Total Favorable Outcomes

The total number of favorable outcomes is the sum of the outcomes from Case 1 and Case 2:

Total Favorable Outcomes = (Number of ways for 3 heads) + (Number of ways for 4 heads)

Total Favorable Outcomes $= 4 + 1 = 5$.

Calculating the Probability

The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes:

$$\text{Probability (Event)} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$$

Substituting the values we found:

• Number of Favorable Outcomes = 5
• Total Number of Possible Outcomes = 16

$$\text{Probability} = \frac{5}{16}$$

Summary of Outcomes and Favorable Events

As shown in the table, the total number of combinations where heads show up more than tails is $4 + 1 = 5$.

Therefore, the probability of this event is $\frac{5}{16}$.