Numbers are selected at random one at a time, from the numbers 00, 01,

The probability of selecting any number pair from 00 to 99 is 1/100. However, there are four possible number pairs whose product result in 18, they are {29, 36, 63, and 92}.
The probability of selecting one of these number pairs in an event is 4/100 = 1/25.
The probability of not selecting one of these number pairs is 1-1/25 = 24/25.
Probability of selecting these number pairs at least three times = probability of selecting these numbers in all four events + probability of selecting these numbers in three out of the events
Probability of selecting these numbers in three out of the four events = 1/25*1/25*1/25*24/25 + 1/25*1/25*24/25*1/25 + 1/25*24/25*1/25*1/25 + 24/25*1/25*1/25*1/25 = (4*24)/(25^4) = 96/390,625
Probability of selecting these numbers in all four events = 1/25*1/25*1/25*1/25 = 1/390,625
P(E)= 1/390,625 + 96/390,625 = 97/390,625

The answer is therefore A.

factors of 18 are 18-1, 9-2, 6-3;
product two digits that make 18 are 9-2, 6-3, 2-9, 3-6, four total;
numbers from 00 to 99 are 99-0+1=100
prob of E is 4 out of 100 = 1/25
prob of at least 3 is probability of EEEN + EEEE
prob EEEN is 4!/3! * (1/25)(1/25)(1/25)(24/25) = 4*24/25^4
prob EEEE is 4!/4! * (1/25)^4 = 1/25^4
total probability is 97/25^4
25*25=625*625=390…

Ans (A)

Numbers are selected at random one at a time, from the numbers 00, 01, 02,…., 99 with replacement. An event E occurs if and only the product of the two digits of a selected number is 18. If four numbers are selected, then the probability that E occurs at least 3 times is

A. 97/390625
B. 98/390625
C. 97/380626
D. 97/380625
E. 98/380625

18 = 1*18 = 2*9 = 3*6
Only last two cases satisfy condition.
Then possible numbers are 29, 92, 36, and 63 i.e. total four numbers are there.

Case I: 3Y's and 1N (where Y is for positive happening of event E and N is negative happening)
Case II: 4Y's

In case I 3 occurrences of Y happen in 4C3 ways i.e. 4 ways out of 100 numbers.
Each positive happening of event E has probability = \(\frac{4}{100}\)

Hence required probability = 1st Number(Y) * 2nd Number(Y) * 3rd Number(Y) * 4th Number(N) + 1st Number(Y) * 2nd Number(Y) * 3rd Number(Y) * 4th Number(Y)
Required Probability = \(\frac{4}{100} * \frac{4}{100} * \frac{4}{100} * \frac{96}{100} * 4 + \frac{4}{100} * \frac{4}{100} * \frac{4}{100} * \frac{4}{100}\)
= \(\frac{4^4 }{ 100^4} (96 + 1)\)
= \(\frac{97}{ 25^4}\) = \(\frac{97 }{ 390625}\)

Answer A.

Numbers are selected at random one at a time, from the numbers 00, 01, 02,…., 99 with replacement. An event E occurs if and only the product of the two digits of a selected number is 18. If four numbers are selected, then the probability that E occurs at least 3 times is

--> 0,1,2,3.....99 - (100 numbers)
The number of an event E -- 29, 92, 36, 63 (4 numbers)

At least 3 times: (E)*(E)*(E)*(not E)
4 ways to select \((\frac{4!}{3!*1!}=4 )\)

--> \(4* (\frac{4}{100})(\frac{4}{100})(\frac{4}{100})(\frac{96}{100})= \frac{96}{5^{8}}\)

4 times: (E)*(E)*(E)*(E)
(just 1 way to select)

--> \((\frac{4}{100})(\frac{4}{100})(\frac{4}{100})(\frac{4}{100})= \frac{1}{5^{8}}\)

Total Probability -- \(\frac{96}{5^{8}} + \frac{1}{5^{8}}= \frac{97}{5^{8}}\)

Answer (A).

The two-digit numbers for which the product of the 2 digits is 18 are: 29, 92, 36, and 63. Therefore, the probability of selecting such a number is 4/100 = 1/25. The probability that such a number occurs at least 3 times is the probability that such a number occurs exactly 3 times plus the probability that such a number occurs all 4 times.

The former probability is 4C3 x (1/25)^3 x (24/25) = 4 x 1/25^3 x 24/25 = 96/25^4.

The latter probability is (1/25)^4 = 1/25^4.

Therefore, the sum of these two probabilities is 96/25^4 + 1/25^4 = 97/25^4 = 97/390,625.

Answer: A

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