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Question

Mathematics Question on Conditional Probability

A die is thrown four times. The probability of getting perfect square in at least one throw is

A

1681\frac{16}{81}

B

6581\frac{65}{81}

C

2381\frac{23}{81}

D

5881\frac{58}{81}

Answer

6581\frac{65}{81}

Explanation

Solution

A die is thrown four time
n=4\therefore n=4
pp (probability of getting perfect square i.e. 1,4))
Here,p=26=13p=\frac{2}{6}=\frac{1}{3}
q=1p=113=23q=1-p=1-\frac{1}{3}=\frac{2}{3}
\therefore Required probability
=P(X1)=1P(X=0)=P(X \geq 1)=1-P(X=0)
=14C0(13)0(23)4=1-{ }^{4} C_{0}\left(\frac{1}{3}\right)^{0}\left(\frac{2}{3}\right)^{4}
=11681=6581=1-\frac{16}{81}=\frac{65}{81}