RRB NTPC/ASM/CA/TA Exam > RRB NTPC/ASM/CA/TA Notes > Mathematics for RRB NTPC (Hindi) > संभावना: हल किए गए उदाहरण- 2
संभावना: हल किए गए उदाहरण- 2 | Mathematics for RRB NTPC (Hindi) - RRB NTPC/ASM/CA/TA PDF Download
उदाहरण 1: एक घड़े में 10 सफेद और 15 काले गेंदें हैं। दो गेंदें बिना प्रतिस्थापन के एक के बाद एक निकाली जाती हैं। संभावना क्या है कि पहली गेंद सफेद है और दूसरी काली है?
समाधान: मान लें A = पहली ड्रॉ में सफेद गेंद मिलना, B = दूसरी ड्रॉ में काली गेंद मिलना। आवश्यक संभावना = पहली ड्रॉ में सफेद गेंद और दूसरी ड्रॉ में काली गेंद मिलने की संभावना = P(A और B) = P(A ∩ B) = P(A) . P(B/A)
P(A) = (10C1/25C1) = (10/25) = (2/5) और P(B/A) = दूसरी ड्रॉ में काली गेंद मिलने की संभावना जब पहली ड्रॉ में सफेद गेंद पहले से है = (15C1/24C1) = (15/24) = (5/8) ⇒ आवश्यक संभावना
उदाहरण 3: हुनर ने CAT पेपर के दो सेक्शन; वर्बल और QA को एक ही क्रम में लिखा। दोनों सेक्शन पास करने की संभावना 0.6 है। वर्बल सेक्शन पास करने की संभावना 0.8 है। वर्बल सेक्शन पास करने के बाद QA सेक्शन पास करने की संभावना क्या है?
कुल संभावना के नियम का उपयोग करते हुए P(गोल्डन क्रिस्टल) = P(A) = P(E1) .
फिर कुल संभावना के नियम द्वारा: P(M) = P(M|L) . P(L) + P(M|NL) . P(NL)
= (2/7).(1/4) + (1/7).(3/4) = 5/28
फिर कुल संभावना के नियम द्वारा:
P(A) = P(A | X)P(X) + P(A | Y)P(Y) + P(A | Z)P(Z).
= (4/10).(1/3) + (1/6) . (1/3) + (3/8) . (1/3) = 113/360 .
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Oct 18, 2025
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