MHT CET · Maths · Probability
Two unbiased dice are thrown. Then the probability that neither a doublet nor a total of 10 will appear is
- A \(\frac{1}{12}\)
- B \(\frac{1}{36}\)
- C \(\frac{2}{9}\)
- D \(\frac{7}{9}\)
Answer & Solution
Correct Answer
(D) \(\frac{7}{9}\)
Step-by-step Solution
Detailed explanation
Number of ways of getting doublet \(=6\)
Number of ways getting a total of 10 are \(\Rightarrow(4,6),(5,5),(6,4)\) i.e. 3 ways
Here \((5,5)\) is common.
\(\therefore\) Total ways of getting doublet or total of 10 are \(6+3-1=8\)
Hence required probability \(=\frac{36-8}{36}=\frac{28}{36}=\frac{7}{9}\)
Number of ways getting a total of 10 are \(\Rightarrow(4,6),(5,5),(6,4)\) i.e. 3 ways
Here \((5,5)\) is common.
\(\therefore\) Total ways of getting doublet or total of 10 are \(6+3-1=8\)
Hence required probability \(=\frac{36-8}{36}=\frac{28}{36}=\frac{7}{9}\)
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