SATHEE: Probability 5 Question 14

Probability 5 Question 14

14. Numbers are selected at random, one at a time, from the two-digit numbers $00, 01, 02, \dots, 99$ 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, $p = P (E) = \frac{4}{100}$

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

$\begin{aligned} P & =^{4} C_{3} p^{3} q +^{4} C_{4} p^{4} \\ = 4 {\frac{4}{100}}^{3} \frac{96}{100} + \frac{4}{100} = \frac{97}{25^{4}} \end{aligned}$

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Probability 5 Question 14

14. Numbers are selected at random, one at a time, from the two-digit numbers 00,01,02,…,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.

Answer:

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