Carcass wrote:

If an integer is randomly selected from all positive 2-digit integers, what is the probability that the integer chosen has

(a) a 4 in the tens place?

(b) at least one 4 in the tens place or the units place?

(c) no 4 in either place?

\((a) \frac{1}{9} (b) \frac{1}{5} (c) \frac{4}{5}\)

Math Review

Question: 11

Page: 297

Difficulty: medium

Here,

a) total number of 2 digit term =(99 - 10) + 1 = 90

Now number of 2 digit that have 4 in tens place = 40, 41 ,42 ......49

Hence there are 10 numbers with 4 in tens place.

Therefore probability a 4 in the tens place = \(\frac{10}{90}= \frac{1}{9}\)

b) at least one 4 in the tens place or the units place

From previous solution, we know probability of 4 in the tens place = 1/9 = Event A

Now,

Number of 4 in units place = 14, 24, 34.....94 = 9 possible terms

probability of 4 in units place = \(\frac{9}{90} = \frac{1}{10}\)= Event B

Using OR probability

We have

P(A or B) = P(A) + P(B) - P(A and B)

or \(P(\frac{1}{9}or \frac{1}{10}) = (\frac{1}{9} + \frac{1}{10})- \frac{1}{90} = \frac{1}{5}\)

C) Now probability of no 4 in either place ( using complement rule) = 1 - ( probability of 4 in either place) = \(1 -\frac{1}{5} = \frac{4}{5}\)

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