Probability Ques 125
Numbers are selected at random, one at a time, from the two-digit numbers $00,01,02, \ldots, 99$ with replacement. An event $E$ occurs if and only if the product of the two digits of a selected number is $18$. If four numbers are selected, find probability that the event $E$ occurs at least $3$ times.
$(1993,5$ M)
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Answer:
Correct Answer: 125.$(\frac{97}{25^{4}})$
Solution:
Formula:
- Let $E$ be the event that product of the two digits is 18 , therefore required numbers are $29,36,63$ and 92 .
Hence, $\quad p=P(E)=\frac{4}{100}$
and probability of non-occurrence of $E$ is
$ q=1-P(E)=1-\frac{4}{100}=\frac{96}{100} $
Out of the four numbers selected, the probability that the event $E$ occurs atleast 3 times, is given as
$ \begin{aligned} P & ={ }^{4} C _3 p^{3} q+{ }^{4} C _4 p^{4} \\ & =4 (\frac{4}{100})^{3} \quad (\frac{96}{100})+(\frac{4}{100} )\quad=\frac{97}{25^{4}} \end{aligned} $
BETA
