A box contains 20% defective bulbs. Five bulb | vedclass

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Question

(A) Let $X$ be the number of defective bulbs chosen. This follows a binomial distribution $B(n, p)$ where $n = 5$ and $p = 20\% = \frac{1}{5}$.Then $q

Vedclass · Accepted Answer

$\frac{32}{625}$

Vedclass · Answer

(A) Let $X$ be the number of defective bulbs chosen. This follows a binomial distribution $B(n, p)$ where $n = 5$ and $p = 20\% = \frac{1}{5}$.Then $q = 1 - p = \frac{4}{5}$.The probability of getting exactly $k$ defective bulbs is given by $P(X = k) = { }^nC_k \cdot p^k \cdot q^{n-k}$.For $k = 3$,we have:$P(X = 3) = { }^5C_3 \cdot (\frac{1}{5})^3 \cdot (\frac{4}{5})^{5-3}$$P(X = 3) = \frac{5 	imes 4}{2 	imes 1} \cdot (\frac{1}{125}) \cdot (\frac{16}{25})$$P(X = 3) = 10 \cdot \frac{16}{3125}$$P(X = 3) = \frac{160}{3125} = \frac{32}{625}$.

$A$ box contains $20\%$ defective bulbs. Five bulbs are chosen randomly from this box. The probability that exactly $3$ of the chosen bulbs are defective is

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