AP EAMCET · Maths · Probability
An unbiased dice with faces marked 1, 2,3, 4,5 , and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5 is
- A \(16 / 81\)
- B \(1 / 81\)
- C \(80 / 81\)
- D \(65 / 81\)
Answer & Solution
Correct Answer
(A) \(16 / 81\)
Step-by-step Solution
Detailed explanation
Here, the face value should be 2, 3, 4, 5 . \(P\) (getting a number not less than 2 and not more than 5 in a single throw) \(=\frac{4}{6}=\frac{2}{3}\) \(\therefore\) Required probability \(=\left(\frac{2}{3}\right)^4=\frac{16}{81}\)
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