Question

Five fair coins are tossed independently. What is the probability that at least two heads appear?

• A. $\frac{13}{16}$
• B. $\frac{7}{16}$
• C. $\frac{5}{16}$
• D. $\frac{11}{16}$

Hint:

For "at least" probability questions, always consider the complement.
Calculating $1 - [P(0) + P(1)]$ requires only two quick binomial computations, whereas direct calculation of $P(2) + P(3) + P(4) + P(5)$ requires four, saving valuable time.

Answer 1

The Correct Option is A

Step 1 : Understanding the Question:
The question asks us to calculate the probability of getting at least two heads when five fair coins are tossed independently.

Step 2 : Key Formulas and Approach:
The number of heads ($X$) obtained in $N$ independent coin tosses follows a Binomial Distribution:
\[ P(X = r) = \binom{N}{r} p^r (1-p)^{N-r} \]
Here, $N = 5$ and $p = \frac{1}{2}$ (for a fair coin).
Using the complement rule is much faster:
\[ P(X \ge 2) = 1 - P(X < 2) = 1 - [P(X = 0) + P(X = 1)] \]

Step 3 : Detailed Explanation:
Let us calculate the individual probabilities for the complementary events:

• Probability of 0 heads ($P(X = 0)$):
\[ P(X = 0) = \binom{5}{0} \left(\frac{1}{2}\right)^5 = 1 \times \frac{1}{32} = \frac{1}{32} \]

• Probability of 1 head ($P(X = 1)$):
\[ P(X = 1) = \binom{5}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^4 = 5 \times \frac{1}{32} = \frac{5}{32} \]
Now, sum the probabilities of the complementary events:
\[ P(X < 2) = P(X = 0) + P(X = 1) = \frac{1}{32} + \frac{5}{32} = \frac{6}{32} = \frac{3}{16} \]
Apply the complement formula to find the probability of getting at least 2 heads:
\[ P(X \ge 2) = 1 - P(X < 2) = 1 - \frac{3}{16} = \frac{13}{16} \]

Step 4 : Final Answer:
The probability that at least two heads appear is $\frac{13}{16}$.
This corresponds to Option (A).