दोन फासे एकाच वेळी फेकले जातात आणि त्यावर दिसणार्या संख्यांची बेरीज नोंदवली जाते. बेरीज 12 असण्याची संभाव्यता किती आहे?

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  1. \(\frac{1}{36}\)
  2. 3
  3. 36
  4. \(\frac{12}{36}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{1}{36}\)
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Detailed Solution

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दिलेल्याप्रमाणे:

दोन फासे एकाच वेळी टाकल्यावर मिळणाऱ्या संभाव्य परिणामांची संख्या:

6 × 6 = 36 

 

(1,1), (1, 2), (1, 3),  (1, 4), (1, 5), (1, 6) 

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) 

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) 

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

वापरलेले सूत्र:

संभाव्यता= अनुकूल परिणामांची संख्या ÷ एकूण परिणामांची संख्या 

गणना:

12 बेरीज मिळणाऱ्या परिणामांची संख्या  (6, 6) = 1

आवश्यक संभाव्यता = 1/36 

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