MHT CET · Maths · Probability
A die is thrown four times. The probability of getting perfect square in at least one throw is
- A \(\frac{58}{61}\)
- B \(\frac{16}{81}\)
- C \(\frac{65}{81}\)
- D \(\frac{23}{81}\)
Answer & Solution
Correct Answer
(C) \(\frac{65}{81}\)
Step-by-step Solution
Detailed explanation
From 1 to 6 , we have 1 and 4 as perfect squares.
Probability of getting perfect square is one throw of a die
\(=\frac{2}{6}=\frac{1}{3}\)
\(\therefore\) Probability of not getting perfect square in 4 throws of a die
\(=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}=\frac{16}{81}\)
\(\therefore\) Required probability \(=1-\frac{16}{81}=\frac{65}{81}\)
Probability of getting perfect square is one throw of a die
\(=\frac{2}{6}=\frac{1}{3}\)
\(\therefore\) Probability of not getting perfect square in 4 throws of a die
\(=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}=\frac{16}{81}\)
\(\therefore\) Required probability \(=1-\frac{16}{81}=\frac{65}{81}\)
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