Question

If 𝑋 is a binomial random variable with 𝑛=16 trials and 𝑝=0.35 probability of success on a given trial, find the probability that we will see 1, 2, or 3 successes.

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Given:

n = 16

p = 0.35

P(X = k) = \binom{16}{k} (0.35)^k (0.65)^{16-k}

Calculate \( P(X = 1) \):

\binom{16}{1}(0.35)^1(0.65)^{15}=16\cdot0.35\cdot(0.65)^{15}\approx0.0087

Calculate \( P(X = 2) \):

\binom{16}{2}(0.35)^2(0.65)^{14}=\frac{16 \cdot15}{2}\cdot(0.35)^2\cdot(0.65)^{14}\approx0.0353

Calculate \( P(X = 3) \):

\binom{16}{3}(0.35)^3(0.65)^{13}=\frac{16 \cdot15 \cdot14}{6}\cdot(0.35)^3\cdot(0.65)^{13}\approx0.0190

Sum the probabilities:

P(X=1)+P(X=2)+P(X=3)=0.063

Therefore, the probability of seeing 1, 2, or 3 successes is:

\boxed{0.063}

Calculate \( P(X = 1) \):

Calculate \( P(X = 2) \):

Calculate \( P(X = 3) \):

Sum the probabilities:

Therefore, the probability of seeing 1, 2, or 3 successes is:

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