Probability Ques 55
An unbiased die 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) $65 / 81$
(1993, 1M)
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Answer:
Correct Answer: 55.(a)
Solution:
Formula:
- Let $A=$ getting not less than $2$ and not greater than $5$
$ \Rightarrow \quad A=\{2,3,4,5\} \quad \Rightarrow P(A)=\frac{4}{6} $
But die is rolled four times, therefore the probability in getting four throws
$ =\begin{array}{llll} (\frac{4}{6}) & (\frac{4}{6}) & (\frac{4}{6}) & (\frac{4}{6}) \end{array}=\frac{16}{81} $
BETA
