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If an integer is randomly selected from all positive 2-digit [#permalink]
Expert's post
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?

[Reveal] Spoiler: OA
$$(a) \frac{1}{9} (b) \frac{1}{5} (c) \frac{4}{5}$$

Math Review
Question: 11
Page: 297
Difficulty: medium

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Re: If an integer is randomly selected from all positive 2-digit [#permalink]
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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?

[Reveal] Spoiler: OA
$$(a) \frac{1}{9} (b) \frac{1}{5} (c) \frac{4}{5}$$

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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Re: If an integer is randomly selected from all positive 2-digit [#permalink]
good question Re: If an integer is randomly selected from all positive 2-digit   [#permalink] 02 Jun 2019, 02:07
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