Visual SolutionPYQ 2024 · Apr Shift 1Standard

Question

A coin is biased with

P(H) = \frac{3}{5}

. It is tossed 4 times. The probability of getting exactly 2 heads is:

(A)

\frac{216}{625}

(B)

\frac{54}{625}

(C)

\frac{12}{25}

(D)

\frac{96}{625}

Solution Path

Binomial:

P(X=2) = \binom{4}{2} (3/5)^2 (2/5)^2 = 6 \cdot 36/625 = 216/625

01Question Setup

1/4

Find

P(X = 2)

for a binomial random variable.

P(X = 2) = \;?

02Binomial Setup

2/4

n=4

p=\frac{3}{5}

q=\frac{2}{5}

k=2

. Binomial formula:

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

P(X=2) = \binom{4}{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^2

03ComputeKEY INSIGHT

3/4

P(X=2) = 6 \cdot \frac{9}{25} \cdot \frac{4}{25} = \frac{216}{625}

P(X=2) = \frac{216}{625}

04Final Answer

4/4

The binomial probability evaluates to

\frac{216}{625}

\boxed{\dfrac{216}{625}}

Concepts from this question1 concepts unlocked

★

EASY

P(A ∩ B) = P(A) × P(B|A). For sequential events, multiply the probability of each stage given all previous outcomes.

P(A \cap B) = P(A) \cdot P(B \mid A)

The backbone of without-replacement and multi-stage probability. Appears in nearly every JEE probability question involving sequential draws.

Without replacementSequential drawsMulti-stage experiments

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