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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Probability

BITSAT 2023 · Paper 1 · Mathematics · Probability

The probability of getting sum more than 7 when a pair of dice are...

The probability of getting sum more than 7 when a pair of dice are thrown is:

Options

  1. A736\frac{7}{36}
  2. B512\frac{5}{12}
  3. C712\frac{7}{12}
  4. DNone of these

Correct answer

B. 512\frac{5}{12}

Step-by-step solution

Here n(S)=62=36n(S)=6^2=36
Let E\mathrm{E} be the event "getting sum more than 7" i.e. sum of pair of dice =8,9,10,11,12=8,9,10,11,12
i.e. E=\left\{(2,6)(3,5)(4,4)(5,3)(6,2)(3,6)(4,5)(5,4)(6,3)(4,6)(5,5)(6,4)(5,6)(6,5)(6,6)\begin{array}{lllll}(2,6) & (3,5) & (4,4) & (5,3) & (6,2) \\ (3,6) & (4,5) & (5,4) & (6,3) \\ (4,6) & (5,5) & (6,4) & \\ (5,6) & (6,5) & (6,6) & \end{array}\right\}
∴n(E)=15.∴\therefore \quad n(E)=15. \therefore Req. prob =n(E)n(S)=1536=512=\frac{n(E)}{n(S)}=\frac{15}{36}=\frac{5}{12}
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