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mreijepatmat


Sample space = {00,10,11,...99} = 100 Total outcomes

from 00 to 99 you can form 100 numbers with 2 digits (including 00).
So any 2 digit number could be written T (tens) and U (unit)  → TU

T can take any value from 0 to 9, then there are 10 ways to choose T
U can take any value from 0 to 9 EXCEPT 5, then there are 9 ways
to choose U.
Then we can select TU in 10 x 9 = 90 ways
Favorable outcome ={90}
P(2 digits NOT ending in 5) = 90/100 = 0.9

The probability that a random 2 digit number (00-99) does not end in 5 is 0.9.

What is probability?

"Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one".

For the given situation,

The event is a random 2 digit number (00-99) does not end in 5.

Let us consider,

In tens place, we can place numbers from 0-9 = 10 numbers.

In units place, we can place numbers from 0-9 but not 5 = 9 numbers.

Total possibilities = 10 × 10 = 100 ways

Thus probability that a random 2 digit number (00-99) does not end in 5 is [tex]P(e) = \frac{(10)(9)}{100}[/tex]

⇒[tex]P(e)=\frac{90}{100}[/tex]

⇒[tex]P(e)=0.9[/tex]

Hence we can conclude that the probability that a random 2 digit number (00-99) does not end in 5 is 0.9.

Learn more about probability here

https://brainly.com/question/13604758

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