Question:medium

If a fair coin is tossed 5 times, what is the probability of getting exactly 3 heads?

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For fair coins ($p=q=1/2$), the formula simplifies to $\frac{\binom{n}{r}}{2^n}$. Just calculate the combination and divide by the total power of 2!
Updated On: Apr 16, 2026
  • \( \frac{5}{32} \)
  • \( \frac{10}{32} \)
  • \( \frac{15}{32} \)
  • \( \frac{20}{32} \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
When a coin is tossed multiple times, the outcomes follow a binomial distribution because there are exactly two possible outcomes (success/head or failure/tail) for each independent toss.
Step 2: Key Formula or Approach:
The binomial probability formula is: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \] where: $n$ = total number of trials
$k$ = number of successful trials
$p$ = probability of success on a single trial
$q$ = probability of failure on a single trial ($1 - p$)
Step 3: Detailed Explanation:
Given a fair coin tossed 5 times: Number of trials, $n = 5$. Desired number of heads, $k = 3$. Probability of getting a head (success), $p = \frac{1}{2}$. Probability of getting a tail (failure), $q = 1 - \frac{1}{2} = \frac{1}{2}$. Plugging these values into the formula: \[ P(X = 3) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{5-3} \] Calculate combinations $\binom{5}{3}$: \[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \] Calculate the powers: \[ P(X = 3) = 10 \times \left(\frac{1}{2}\right)^3 \times \left(\frac{1}{2}\right)^2 = 10 \times \left(\frac{1}{2}\right)^5 = 10 \times \frac{1}{32} = \frac{10}{32} \]
Step 4: Final Answer:
The probability of getting exactly 3 heads is $\frac{10}{32}$.
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