Question

Four unbiased coins are tossed simultaneously. Probability of getting at most two heads is ________.

• A. $\frac{5}{8}$
• B. $\frac{9}{16}$
• C. $\frac{11}{16}$
• D. $\frac{13}{16}$
• E. $\frac{15}{16}$

Hint:

"At most 2" means 0, 1, or 2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
This is a problem of binomial probability. We are tossing a fair coin 4 times, which is a sequence of 4 independent Bernoulli trials. "Atmost two heads" means we can get 0 heads, 1 head, or 2 heads. We need to calculate the probability of each of these events and add them up.
Step 2: Key Formula or Approach
The probability of getting exactly \(k\) successes in \(n\) independent Bernoulli trials is given by the binomial probability formula:
\[ P(X=k) = {}^nC_k \cdot p^k \cdot (1-p)^{n-k} \] Here, \(n=4\) (number of tosses), \(p = 0.5\) (probability of getting a head), and \(1-p = 0.5\) (probability of getting a tail).
We need to find \(P(\text{atmost 2 heads}) = P(X=0) + P(X=1) + P(X=2)\).
The total number of possible outcomes is \(2^4 = 16\).
Step 3: Detailed Explanation
1. Calculate the total number of outcomes.
Each of the 4 coins can land in 2 ways (Heads or Tails). So, the total number of possible outcomes in the sample space is \(2 \times 2 \times 2 \times 2 = 2^4 = 16\).
2. Calculate the number of favorable outcomes.
"Atmost two heads" means the number of heads (k) can be 0, 1, or 2.
- Case 1: 0 Heads (k=0)
This means all are tails (TTTT). The number of ways to get 0 heads is \({}^4C_0 = \frac{4!}{0!4!} = 1\).
- Case 2: 1 Head (k=1)
This means one head and three tails (e.g., HTTT). The number of ways to arrange this is \({}^4C_1 = \frac{4!}{1!3!} = 4\).
- Case 3: 2 Heads (k=2)
This means two heads and two tails (e.g., HHTT). The number of ways to arrange this is \({}^4C_2 = \frac{4!}{2!2!} = \frac{24}{4} = 6\).
3. Sum the favorable outcomes.
Total number of favorable outcomes = (Ways for 0 heads) + (Ways for 1 head) + (Ways for 2 heads)
Total favorable outcomes = \(1 + 4 + 6 = 11\).
4. Calculate the probability.
\[ P(\text{atmost 2 heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{11}{16} \] Step 4: Final Answer
The probability of getting atmost two heads is \(\frac{11}{16}\).