यदि स्वतंत्र यादृच्छिक चर X,Y को द्विपद रूप से क्रमशः n = 3, p = \(\dfrac{1}{3}\) और n = 5, p = \(\dfrac{1}{3}\) के साथ बंटन किया जाता है, तो (X + Y ≥ 1) की प्रायिकता है:

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SSC CGL JSO Tier-II, 2018 (Statistics) Official Paper-III (Held On: 14 Sept, 2019)
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  1. 1 - \((\dfrac{2}{3})^6\)
  2. 1 - \((\dfrac{1}{3})^8\)
  3. 1 - \((\dfrac{2}{3})^8\)
  4. 1 - \((\dfrac{1}{3})^6\)

Answer (Detailed Solution Below)

Option 3 : 1 - \((\dfrac{2}{3})^8\)
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दिया गया है:

यदि स्वतंत्र यादृच्छिक चर X,Y को द्विपद रूप से बंटन किया जाता है तो इसका अर्थ है

~ B(n, p)

~ B(3, 1/3) और Y ~ B(5, 1/3)

यहां, B = द्विपद बंटन

n = प्रेक्षणों की संख्या

p = प्रायिकता

गणना:

चूंकि X और Y स्वतंत्र द्विपद यादृच्छिक चर हैं जिनमें p1 = p2 है = 1/3

द्विपद बंटन का गुण जोड़ने पर

⇒ X + Y ~ B(3 + 5, 1/3)

⇒ X + Y ~B(8,1/3)

हम जानते हैं कि, P(X + Y ≥ γ) = 8Cr(1/3)r(2/3)8 - r

⇒ P(X + Y ≥ 1) = 1 - P( X + Y < 1)

⇒ 1 - P(X + Y = 0)

∴ 1 - (2/3)8

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