Question

A fair coin is tossed 4 times. If $X$ is a random variable which indicates number of heads, then $P[X \lt 3] =$

• A. $\frac{10}{16}$
• B. $\frac{1}{16}$
• C. $\frac{12}{16}$
• D. $\frac{11}{16}$

Hint:

For 4 coin tosses, the total number of outcomes in the sample space is $2^4 = 16$. The counts of outcomes for $X = 0, 1, 2, 3, 4$ follow the symmetric row of Pascal's Triangle: 1, 4, 6, 4, 1. Adding the counts for $X=0, 1, 2$ gives $1 + 4 + 6 = 11$, directly yielding $\frac{11}{16}$ without using formulas!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A fair coin is flipped $n = 4$ times. We define a discrete random variable $X$ representing the total number of heads obtained. We need to calculate the cumulative probability that we get strictly fewer than 3 heads, written as $P[X \lt 3]$.

Step 2: Key Formula or Approach:
Coin flipping follows a Bernoulli trial system modeled by the Binomial Distribution $X \sim B(n, p)$, where $n = 4$ and the probability of getting a head on a single flip is $p = \frac{1}{2}$ (so $q = 1 - p = \frac{1}{2}$). The probability mass formula is: $$P(X = k) = \binom{n}{k} p^k q^{n-k}$$ The condition $X \lt 3$ means $X$ can take the values 0, 1, or 2. Alternatively, using the complement rule: $$P[X \lt 3] = 1 - (P[X = 3] + P[X = 4])$$ Computing the complement is faster since it involves fewer terms.

Step 3: Detailed Explanation:
Let's apply the complement method to solve the problem: $$P[X \lt 3] = 1 - \left( P(X=3) + P(X=4) \right)$$ Calculate the individual probability values using our binomial formula: 1. For $k = 3$: $$P(X = 3) = \binom{4}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{4-3} = 4 \times \frac{1}{8} \times \frac{1}{2} = \frac{4}{16}$$ 2. For $k = 4$: $$P(X = 4) = \binom{4}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^{4-4} = 1 \times \frac{1}{16} \times 1 = \frac{1}{16}$$ Now, substitute these values back into our complement equation: $$P[X \lt 3] = 1 - \left( \frac{4}{16} + \frac{1}{16} \right) = 1 - \frac{5}{16}$$ Convert 1 to a fraction with a common denominator of 16: $$P[X \lt 3] = \frac{16}{16} - \frac{5}{16} = \frac{11}{16}$$

Step 4: Final Answer:
The probability of getting fewer than 3 heads is $\frac{11}{16}$, which corresponds to option (D).