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Question

(A) Let $G$ denote a good article and $B$ denote a bad article. Total articles $= n + m$.We are picking $2$ articles in succession without replacement

Vedclass · Accepted Answer

$\frac{m}{m+n}$

Vedclass · Answer

(A) Let $G$ denote a good article and $B$ denote a bad article. Total articles $= n + m$.We are picking $2$ articles in succession without replacement.The second article is bad in two mutually exclusive cases:$1$. The first article is bad and the second article is bad $(B_1 \cap B_2)$.$2$. The first article is good and the second article is bad $(G_1 \cap B_2)$.The probability is given by:$P(B_2) = P(B_1 \cap B_2) + P(G_1 \cap B_2)$$P(B_2) = P(B_1) \cdot P(B_2|B_1) + P(G_1) \cdot P(B_2|G_1)$$P(B_2) = \left( \frac{m}{n+m} \right) \cdot \left( \frac{m-1}{n+m-1} \right) + \left( \frac{n}{n+m} \right) \cdot \left( \frac{m}{n+m-1} \right)$$P(B_2) = \frac{m(m-1) + nm}{(n+m)(n+m-1)}$$P(B_2) = \frac{m^2 - m + nm}{(n+m)(n+m-1)}$$P(B_2) = \frac{m(m + n - 1)}{(n+m)(n+m-1)}$$P(B_2) = \frac{m}{n+m}$

From a lot containing $n$ good and $m$ bad articles, if $2$ articles are picked at random in succession without replacement, then the probability that the second article picked is bad is

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