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Let x ~ Binomial (5,0.6) and Y ~ Poisson (2) be independent. Then P(xy = 0) equals:

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SSC CGL Tier-II ( JSO ) 2016 Official Paper ( Held On : 2 Dec 2016 )
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  1. e-2. (0.4)5
  2. e-2 + (0.4)5
  3. e-2 + (0.4)5 - e-2. (0.4)5
  4. e-2 + (0.6)5 + (1 - e-2) (0.4)5

Answer (Detailed Solution Below)

Option 3 : e-2 + (0.4)5 - e-2. (0.4)5
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Detailed Solution

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Formula

In poisson distribution

P(X = x) = e-λλx/x! ; x = 0, 1, 2, 3, ------ λ > 0

In Binomial distribution

P(x) = ncxpxqλ - x

Calculation

P(xy = 0) = e-2 × 20/0! + 5c0(0.4)5 × (0.6)0

⇒  e-2 × 20/0! + 5c0(0.4)5

⇒ e-2 + (0.4)5 - e-2 × (0.4)5

∴ The P(xy = 0) equals to e-2 + (0.4)5 - e-2 × (0.4)5

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