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Q.A factory has two machines I and II. Machine I produces 40% of items of the output and Machine II produces 60% of the items. Further 4% of items produced by Machine I are defective and 5% produced by Machine II are defective. An item is drawn at random. If the drawn item is defective, find the probability that it was produced by Machine II. OR If y=(cos⁡−1x)2y=(\cos^{-1}x)^2 prove that (1−x2)d2ydx2−xdydx−2=0(1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} - 2 = 0, hence find y2y_2 when x=0x=0.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2023Subjective· 5mImportance★★★★★
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Using Bayes' theorem, the probability that a defective item came from Machine II is 1523\dfrac{15}{23}.

Let II, IIII denote the events "item from Machine I" and "item from Machine II", and DD the event "item is defective".

Given: P(I)=0.40P(I)=0.40, P(II)=0.60P(II)=0.60, P(D∣I)=0.04P(D\mid I)=0.04, P(D∣II)=0.05P(D\mid II)=0.05.

By the law of total probability:

P(D)=P(I)P(D∣I)+P(II)P(D∣II)=(0.40)(0.04)+(0.60)(0.05)=0.016+0.030=0.046P(D) = P(I)P(D\mid I) + P(II)P(D\mid II) = (0.40)(0.04)+(0.60)(0.05) = 0.016+0.030 = 0.046

By Bayes' theorem:

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