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Q.A factory produces bulbs, of which 6% are defective bulbs in a large bulk of bulbs. Based on the above information, answer the following questions :

(i) Find the probability that in a sample of 100 bulbs selected at random, none of the bulbs is defective. (Use : e−6=0⋅0024e^{-6} = 0\cdot0024) [1]
(ii) Find the probability that the sample of 100 bulbs has exactly two defective bulbs. [1]
(iii)
(a) Find the probability that the sample of 100 bulbs will include not more than one defective bulb. [2]
(OR)
(iii)
(b) Find the mean and the variance of the distribution of number of defective bulbs in a sample of 100 bulbs. [2]
CBSECBSE Class XII Board 2023Subjective· 4mImportance★★★★★
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Poisson with λ=6\lambda=6: P(0)=0.0024P(0)=0.0024, P(2)=0.0432P(2)=0.0432, P(X≤1)=0.0168P(X\le1)=0.0168; mean == variance =6=6.

P(X=k)=e−λλkk!P(X=k)=\dfrac{e^{-\lambda}\lambda^{k}}{k!} with λ=np\lambda=np; for a Poisson distribution mean =λ=\lambda and variance =λ.=\lambda.

  1. n=100, p=0.06⇒λ=np=6.n=100,\ p=0.06\Rightarrow \lambda=np=6.
  1. No defective bulb: 2. P(X=0)=e−6600!=e−6=0.0024.P(X=0)=\dfrac{e^{-6}6^{0}}{0!}=e^{-6}=0.0024.
  2. Exactly two defective: 3. P(X=2)=e−6622!=0.0024×362=0.0024×18=0.0432.P(X=2)=\dfrac{e^{-6}6^{2}}{2!}=\dfrac{0.0024\times36}{2}=0.0024\times18=0.0432. (iii)(a) Not more than one defective: …

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