Bernoulli trials — probability of exactly 3 successes in 5 trials

Question

A die is thrown 5 times. If getting a number greater than 4 is considered a “success,” find the probability of getting exactly 3 successes.

(CBSE 2023)

Solution — Step by Step

A “success” is getting 5 or 6 on a die. So:

$p = P(\text{success}) = \frac{2}{6} = \frac{1}{3}$

$q = P(\text{failure}) = 1 - \frac{1}{3} = \frac{2}{3}$

Number of trials: $n = 5$ , Required successes: $k = 3$ .

P(X = k) = \binom{n}{k} p^k q^{n-k}

P(X = 3) = \binom{5}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^2

\binom{5}{3} = 10

\left(\frac{1}{3}\right)^3 = \frac{1}{27}

\left(\frac{2}{3}\right)^2 = \frac{4}{9}

P(X = 3) = 10 \times \frac{1}{27} \times \frac{4}{9} = 10 \times \frac{4}{243} = \frac{40}{243}

The answer is $\frac{40}{243}$ $\approx 0.165$ .

Why This Works

Each die throw is independent, with the same probability of success ( $1/3$ ). This makes it a sequence of Bernoulli trials — repeated independent experiments with exactly two outcomes (success/failure) and constant probability.

The binomial formula counts it this way: $\binom{5}{3}$ is the number of ways to choose which 3 of the 5 throws are successes. For each such arrangement, the probability is $p^3 q^2$ (3 successes at $1/3$ each, 2 failures at $2/3$ each). Multiplying gives the total probability.

The term $\frac{40}{243}$ means roughly a 16.5% chance — getting 3 out of 5 when your success probability is only $1/3$ is not very likely, which matches our intuition.

Alternative Method — Check with complementary approach

We can verify by computing $\sum_{k=0}^{5} P(X = k) = 1$ :

$P(0) = \frac{32}{243}$ , $P(1) = \frac{80}{243}$ , $P(2) = \frac{80}{243}$ , $P(3) = \frac{40}{243}$ , $P(4) = \frac{10}{243}$ , $P(5) = \frac{1}{243}$

Sum: $\frac{32 + 80 + 80 + 40 + 10 + 1}{243} = \frac{243}{243} = 1$ ✓

For CBSE and JEE, always simplify the fraction — don’t leave it as $10 \times 4/243$ . Also, a common question variation: “find the probability of at least 3 successes.” Then you’d compute $P(3) + P(4) + P(5) = \frac{40 + 10 + 1}{243} = \frac{51}{243} = \frac{17}{81}$ . Know the difference between “exactly” and “at least.”

Common Mistake

Students sometimes define $p = 4/6$ instead of $2/6$ , misreading “greater than 4” as “4 or greater.” The phrase “greater than 4” means strictly greater — so only 5 and 6 qualify, giving $p = 2/6 = 1/3$ . “Greater than or equal to 4” would include 4, 5, 6, giving $p = 3/6 = 1/2$ . Read the problem statement carefully.

Question

Solution — Step by Step

Why This Works

Alternative Method — Check with complementary approach

Common Mistake

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