MHT CET · Maths · Probability
Two dice are thrown together. The probability that sum of the numbers is divisible by 2 or 3 is
- A \(\frac{1}{6}\)
- B \(\frac{3}{4}\)
- C \(\frac{1}{3}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
Two dice are thrown together. Then sum of \(2,3,4,6,8,9,10,12\) is obtained in following ways.
Let \(A=\{(2,2),(1,2),(1,1),(1,3),(1,5),(2,1),(2,4),\) \((2,6),(3,1),(3,3),(3,5),(3,6),\) \((4,5),(4,2),(4,4),(4,6),(5,1),\)\((5,3),(5,4),(5,5),(6,2),\) \((6,3),(6,4),(6,6)\}\)
Thus \(n(A)=24\) and \(n(S)=6 \times 6=36\)
\(\therefore \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{24}{36}=\frac{2}{3}\)
Let \(A=\{(2,2),(1,2),(1,1),(1,3),(1,5),(2,1),(2,4),\) \((2,6),(3,1),(3,3),(3,5),(3,6),\) \((4,5),(4,2),(4,4),(4,6),(5,1),\)\((5,3),(5,4),(5,5),(6,2),\) \((6,3),(6,4),(6,6)\}\)
Thus \(n(A)=24\) and \(n(S)=6 \times 6=36\)
\(\therefore \mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{24}{36}=\frac{2}{3}\)
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