MHT CET · Maths · Probability
The odds in favour of getting sum multiple of 3, when pair of dice are thrown is
- A \(4: 5\)
- B \(2: 3\)
- C \(1: 2\)
- D \(3: 4\)
Answer & Solution
Correct Answer
(C) \(1: 2\)
Step-by-step Solution
Detailed explanation
(A)
When a pair of dice is thrown, then total outcomes are \(6 \times 6=36\).
Now the odds in favour of getting the sum, which is multiple of 3 are
\((1,2)(2,1) \Rightarrow \operatorname{sum} 3\)
\((3,3),(2,4),(4,2),(1,5),(5,1) \Rightarrow \operatorname{sum} 6\)
\((4,5),(5,4),(6,3),(3,6) \Rightarrow \operatorname{sum} 9\)
\((6,6) \Rightarrow\) sum 12
Thus Number of favourable cases \(=2+5+4+1=12\)
So, odds in fovour \(=\frac{12}{24}=\frac{1}{2}\)
When a pair of dice is thrown, then total outcomes are \(6 \times 6=36\).
Now the odds in favour of getting the sum, which is multiple of 3 are
\((1,2)(2,1) \Rightarrow \operatorname{sum} 3\)
\((3,3),(2,4),(4,2),(1,5),(5,1) \Rightarrow \operatorname{sum} 6\)
\((4,5),(5,4),(6,3),(3,6) \Rightarrow \operatorname{sum} 9\)
\((6,6) \Rightarrow\) sum 12
Thus Number of favourable cases \(=2+5+4+1=12\)
So, odds in fovour \(=\frac{12}{24}=\frac{1}{2}\)
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