Q.In the same coin-tossing experiment (4 tosses of a fair coin), find the probability of getting at least 3 heads.
Concept understanding — Binomial Distribution
Binomial Distribution: From Intuition to Precision
Imagine you flip a fair coin 10 times. You want to know the probability of getting exactly 6 heads. This is the kind of question the binomial distribution answers — it models the number of "successes" in a fixed number of independent trials, where each trial has only two outcomes.
The Intuition
Think of a single trial. You roll a die and call "getting a 4" a success. That's one trial. Now roll the die 5 times. The number of times you get a 4 could be 0, 1, 2, 3, 4, or 5. Each roll is independent — the result of one roll doesn't affect the next. The probability of success (getting a 4) stays the same every time: p=61.
The binomial distribution tells you exactly how likely each possible count of successes is, given:
- a fixed number of trials n,
- a constant success probability p per trial,
- independent trials.
The Key Conditions
For a situation to be modelled by a binomial distribution, all four must hold:
- Fixed number of trials (n). You decide in advance how many times you'll repeat the experiment.
- Two outcomes per trial — "success" and "failure". These are just labels; success is whatever outcome you're counting.
- Constant probability of success (p) on every trial.
- Independent trials. The outcome of one trial does not influence another.
A common mistake is to apply the binomial distribution when trials are not independent — for example, drawing cards without replacement from a deck. That's a hypergeometric situation, not binomial.
The Precise Statement
Let X be the number of successes in n independent trials, each with success probability p. Then X follows a binomial distribution with parameters n and p, written:
X∼Binomial(n,p)
The probability of getting exactly k successes (where k=0,1,2,…,n) is:
P(X=k)=(kn)pk(1−p)n−k
P(X=k)=(kn)pk(1−p)n−k
Breaking Down the Formula
Three pieces multiply together:
- pk — the probability that the k successes happen. Since each success has probability p, and there are k of them, this factor is p multiplied by itself k times.
- (1−p)n−k — the probability that the remaining n−k trials are failures. Each failure has probability 1−p.
- (kn) — the number of ways to choose which k of the n trials are the successes. The successes could be the first k trials, or the last k, or any arrangement. This binomial coefficient counts all possible arrangements.
The binomial coefficient (kn) is read as "n choose k" and equals k!(n−k)!n!. For example, (25)=2!3!5!=10.
A Worked Example
Problem: A multiple-choice test has 10 questions, each with 4 options. You guess every answer. What is the probability you get exactly 3 correct?
Solution:
- n=10 (10 trials)
- p=41 (probability of guessing correctly on one question)
- k=3 (we want exactly 3 successes)
P(X=3)=(310)(41)3(43)7
Compute (310)=120, so:
P(X=3)=120×641×(43)7
(43)7=163842187, so:
P(X=3)=120×641×163842187=64×16384120×2187
Simplify: 120/64=15/8, so:
P(X=3)=8×1638415×2187=13107232805≈0.2503
So there's about a 25% chance of getting exactly 3 correct by pure guessing.
Mean and Variance
For a binomial distribution, two important summary measures are:
- Mean (expected number of successes): μ=np
- Variance: σ2=np(1−p)
The mean makes intuitive sense: if you flip a coin 100 times with p=0.5, you expect about 50 heads. The variance tells you how spread out the distribution is — it's largest when p=0.5 (maximum uncertainty) and smallest when p is near 0 or 1 (outcome is almost certain).
The binomial distribution is the foundation for many statistical tests and is closely related to the normal distribution — for large n, a binomial distribution with np and np(1−p) both at least 5 can be approximated by a normal distribution with the same mean and variance.
When Not to Use It
The binomial distribution fails if any condition is violated. Common exam traps:
- Without replacement from a finite population — use hypergeometric.
- Trials continue until a success — use geometric distribution.
- Counting events in a fixed interval — use Poisson distribution.
The binomial distribution is your tool when you have a fixed number of independent, identical trials, each with two outcomes. Master this, and you've unlocked a core piece of probability.
Binomial distribution is a core topic of the NCERT Class 12 Mathematics chapter on Probability, and searches like "binomial distribution formula and examples" or "binomial distribution class 12 important questions" reflect how frequently it appears in CBSE board papers and JEE Main statistics questions. The mean-variance relationship and the contrast with Poisson and hypergeometric distributions covered above are exactly the distinctions competitive exams like to test.
"At least 3 heads" out of 4 tosses means exactly 3 heads or exactly 4 heads, so the two individual binomial probabilities must be added.
P(X≥3) = P(X=3) + P(X=4) = 0.25 + 0.0625.
P(at least 3 heads) = 0.3125
With n=4, p=q=0.5, "at least 3 heads" is the event X=3 or X=4:
P(X=3)=(34)(0.5)3(0.5)1=4×0.125×0.5=0.25
P(X=4)=(44)(0.5)4(0.5)0=1×0.0625=0.0625
P(X≥3)=P(X=3)+P(X=4)=0.25+0.0625=0.3125
P(at least 3 heads) = 0.3125
As a cross-check, compute the complement instead. The full distribution for n=4, p=0.5 is P(0)=0.0625, P(1)=0.25, P(2)=0.375, P(3)=0.25, P(4)=0.0625 (these sum to 1.0000, confirming the table is correct). Then P(X≥3) = 1 − [P(0)+P(1)+P(2)] = 1 − (0.0625+0.25+0.375) = 1 − 0.6875 = 0.3125, which matches the direct addition exactly.
Students sometimes read "at least 3" as "more than 3" and compute only P(X=4), missing the X=3 term. Always translate "at least" as ≥, including the boundary value.
- CA Foundation 2026Set jan-20261 markMCQQ.A quality control inspector finds that 20% of light bulbs are defective. If a batch of 5 light bulbs is tested, what is the probability that exactly 1 bulb is defective? (A) 0.4096 (B) 0.8026 (C) 0.2746 (D) 0.1296
›Reveal solutionSolution
P(X=1)=(15)(0.2)1(0.8)4=0.4096.
Step 1 — Set up the binomial model
Each bulb is defective with probability p=0.2 independently; n=5 trials.
Step 2 — Apply the binomial formula for X=1
P(X=1)=(15)p1(1−p)4=5×0.2×(0.8)4.
Step 3 — Compute
(0.8)4=0.4096,P=5×0.2×0.4096=1×0.4096=0.4096.
Watch out"Exactly 1 defective" means the other 4 bulbs are good, so the (0.8)4 term is essential — using 0.2 alone (or forgetting the (15) factor) gives the wrong value.
TipNotice 5×0.2=1, so the answer collapses to just (0.8)4=0.4096 — no long multiplication needed.
✓Final answer(A) 0.4096
- CA Foundation 2025Set may-20251 markMCQQ.What is the probability of making 3 corrected guesses in 5 True-False answer type questions ? (A) 0.3125 (B) 0.4156 (C) 1.3888 (D) 0.5235
›Reveal solutionSolution
Binomial with n=5,p=21: P(X=3)=(35)(21)5=3210=0.3125.
Step 1 — Identify the model
Each of 5 True-False questions is an independent trial with p=21 of a correct guess, so X= number correct is Binomial.
P(X=r)=(rn)pr(1−p)n−r
Step 2 — Substitute n=5, r=3, p=21
P(X=3)=(35)(21)3(21)2=10×(21)5
Step 3 — Evaluate
P(X=3)=3210=0.3125
Why the other options are wrong: (C) 1.3888 exceeds 1 (impossible); (B) 0.4156 and (D) 0.5235 use wrong p or coefficient.
Watch outA probability can never exceed 1, so option (C) is instantly wrong. Also remember (21)3(21)2=(21)5, not (21)3.
TipWith p=21, every binomial term is just 2n(rn).
✓Final answer(A) 0.3125
- CA Foundation 2025Set sep-20251 markMCQQ.The Mode of binomial distribution B(7,1/3) is (A) 3 (B) 2 (C) 7/3 (D) 8/3
›Reveal solutionSolution
Mode of B(n,p) = ⌊(n+1)p⌋ when (n+1)p is non-integer; here (8)(1/3)=8/3 → mode = 2.
Step 1 — The mode formula for a binomial
Mode=⌊(n+1)p⌋if (n+1)p is not an integer
Step 2 — Substitute n = 7, p = 1/3
(n+1)p=(7+1)×31=38≈2.67
Since 2.67 is not an integer, take its integer part: mode =2.
Why the other options are wrong: 3 would be ⌈8/3⌉; 7/3 and 8/3 are the mean np=7/3 and the value (n+1)p=8/3 respectively — neither is the mode, which must be a whole-number value the variable can actually take.
Watch outThe mode must be an achievable count (an integer 0…n). Options like 7/3 or 8/3 can be discarded immediately because a binomial variable is discrete.
TipCompute (n+1)p; if it lands between two integers, the floor is the single mode (if it is an integer, that value and the one below it are both modes).
✓Final answer(B) 2
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