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Suppose a discrete random variable can only take the values 0, 1, and 2. The probability mass function is defined by ,for,,,otherwisef(x)={x2+1k, for x=0, 1, 20, otherwise Find P(X ≥ 1)

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प्रश्न

Suppose a discrete random variable can only take the values 0, 1, and 2. The probability mass function is defined by 
`f(x) = {{:((x^2 + 1)/k","  "for"  x = 0","  1","  2),(0","  "otherwise"):}` 
Find P(X ≥ 1)

योग
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उत्तर

P(X ≥ 1) = = P(X = 1) + P(X = 2)

= `2/8 + 5/8 + 7/8`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Probability Distributions - Exercise 11.2 [पृष्ठ १९४]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 11 Probability Distributions
Exercise 11.2 | Q 4. (iii) | पृष्ठ १९४

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