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Probability Ncert Solutions Maths class 12th Maths Class 12th Probability

43. Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability if 0.3, if the second group wins. Find the probability that the new product introduced was by the second group.

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alok kumar singh | Contributor-Level 10

4 months ago

43. Let, $E_{1} :$ event that first group will win

$E_{2} :$ event that second group will win

$A :$ event that new product will produce

$P (E_{1}) = 0.6 = \frac{6}{10} = \frac{3}{5}$

$P (E_{2}) = 0.4 = \frac{4}{10} = \frac{2}{5}$

$P (A / E_{1}) = P$ (introducing new product by group $A$ ) $= 0.7 = \frac{7}{10}$

$P (A / E_{2}) = P$ ( introducing new product by group $B$ ) $= 0.3 = \frac{3}{10}$

Therefore, by Baye’s theorem,

$P (E_{2} / A) =$ probability that new product introduced was produced by second group

$P (E_{2} / A) = \frac{P (E_{2}) . P (A / E_{2})}{P (E_{1}) . P (A / E_{1}) + P (E_{2}) . P (A / E_{2})}$

$= \frac{\frac{2}{5} \times \frac{3}{10}}{\frac{3}{7} \times \frac{7}{10} + \frac{2}{5} \times \frac{3}{10}}$

$= \frac{\frac{6}{50}}{\frac{21}{50} + \frac{6}{50}}$

$= \frac{6}{50} \times \frac{50}{27} = \frac{2}{9}$

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