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Q.It is given that 3% defective electric bulb are manufactured by a company. Using Poisson distribution, find the probability of 100 bulbs will contain no defective bulbs. (Use e−3e^{-3} = 0.05)

Yanam CbseNCERTSubjective· 3mImportance★★★★★
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With defect rate 3%3\% over 100100 bulbs, λ=np=3\lambda=np=3; the probability of no defective bulb is P(0)=e−3=0.05.P(0)=e^{-3}=0.05.

λ=np\lambda=np, P(X=k)=e−λλkk!\quad P(X=k)=\dfrac{e^{-\lambda}\lambda^{k}}{k!}

where nn = number of bulbs, pp = defect probability.

Steps

  1. Given p=3%=0.03p=3\%=0.03 and n=100.n=100.

  2. Poisson parameter: λ=np=100×0.03=3.\lambda=np=100\times0.03=3. …

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