It is currently 17 Nov 2019, 01:31

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# If an integer is randomly selected from all positive 2-digit

Author Message
TAGS:
Founder
Joined: 18 Apr 2015
Posts: 8760
Followers: 175

Kudos [?]: 2034 [0], given: 8089

If an integer is randomly selected from all positive 2-digit [#permalink]  30 May 2019, 16:19
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

_________________

Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos
GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.

VP
Joined: 20 Apr 2016
Posts: 1087
WE: Engineering (Energy and Utilities)
Followers: 16

Kudos [?]: 921 [2] , given: 226

Re: If an integer is randomly selected from all positive 2-digit [#permalink]  01 Jun 2019, 08:16
2
KUDOS
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) = $$1 -\frac{1}{5} = \frac{4}{5}$$
_________________

If you found this post useful, please let me know by pressing the Kudos Button

Rules for Posting

Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months GRE Prep Club tests

Intern
Joined: 26 May 2018
Posts: 37
Followers: 0

Kudos [?]: 7 [0], given: 2

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