Probability 5 Question 14
14. 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: 14. $\frac{97}{25^{4}}$
Solution:
- 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
$$ \begin{aligned} P & ={ }^{4} C _3 p^{3} q+{ }^{4} C _4 p^{4} \\ & =4 \frac{4}{100}^{3} \quad \frac{96}{100}+\frac{4}{100} \quad=\frac{97}{25^{4}} \end{aligned} $$
